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text:
\section{Introduction}

Although linear boundary problems play an important role in
applied mathematics~\cite{Agarwal1986,Agarwal2008,Duffy2001,Stakgold1979},
there is little algebraic theory and algorithmic treatment of boundary problems.
Current computer algebra systems provide many symbolic tools for
differential equations, but boundary conditions are usually left to a backward solving procedure,
which---depending on the forcing function and on the conditions---may or may not work.

In~\cite{Rosenkranz2005}, a new operator based approach for symbolic computation with linear ordinary boundary problems was presented,
which has constantly been extended over the last
years~\cite{RegensburgerRosenkranz2009,RosenkranzRegensburger2008a,RosenkranzRegensburgerTecBuchberger2009};
see also~\cite{RosenkranzRegensburgerTecBuchberger2012} for a recent overview.
The results needed are summarized in Section~\ref{sec:SC}.

The most recent algorithms for regular boundary problems (that are uniquely solvable) are implemented
in the TH$\exists$OREM$\forall$\ system \cite{RosenkranzRegensburgerTecBuchberger2009,Tec2011},
and in the \textsc{Maple}\ package \texttt{IntDiffOp}~\cite{KorporalRegensburgerRosenkranz2011,KorporalRegensburgerRosenkranz2012,Korporal2012}.
They do not only allow to compute solution operators (Green's operators), but also to factor regular boundary problems
into lower order problems provided a factorization of the underlying differential operator.
The factorization of boundary problems relies on the  multiplicative structure introduced in~\cite{RegensburgerRosenkranz2009},
which for regular problems corresponds to the multiplication of the respective solution operators in reverse order.

The \texttt{IntDiffOp}\ package also provides support for the class of singular problems treated in this paper:
We consider boundary problems where the differential equation per se is solvable, but where inconsistent
boundary conditions allow solutions only for forcing functions satisfying suitable \emph{compatibility conditions}. As a simple example, consider the boundary problem
 \begin{equation}
  \label{eq:bpsingular}
  \bvp{u''(x) = f(x)}{u(1)=u'(1)=u'(0)=0,}
 \end{equation}
where the forcing function $f$ clearly has to satisfy the compatibility condition  $\textstyle{\int_0^1} f(\xi)\,\mathrm{d}\xi = 0$; see Example~\ref{ex:CC} for the corresponding code.

While Green's operators for regular boundary problems are right inverses of the differential operator, solution operators for singular problems can be described
as generalized inverses.
Algorithms for computing such generalized Green's operators and the compatibility conditions of a singular boundary problem are
presented in~\cite{KorporalRegensburgerRosenkranz2011}; we briefly recall the basic results in Section~\ref{sec:GGOP}.
For singular boundary problems and
generalized or modified Green's functions in ana\-ly\-sis, we refer for example to~\cite{Stakgold1979}
and~\cite{Loud1970}, and in the context of generalized inverses to
\cite[Sec.~9.4]{Ben-IsraelGreville2003}, \cite{Boichuk2004}, and~\cite[Sec.~H]{NashedRall1976}.

The goal of this paper is to extend the factorization algorithm for regular boundary
problems~\cite{RegensburgerRosenkranz2009,RosenkranzRegensburgerTecBuchberger2009} to generalized
boundary problems. To this end, we have earlier investigated the multiplicative
structure of generalized inverses in~\cite{KorporalRegensburger2013}. It turns out that, in contrast to the regular case, the product of generalized Green's operators is not, in general, itself a
generalized Green's operator.

In Section~\ref{sec:Composition}, we define a new composition of generalized boundary problems,
which includes the composition of regular problems, based on results from~\cite{Korporal2012,KorporalRegensburger2013}.
Then we discuss algorithmic methods for testing when the so-called reverse order law holds, that is, when the product
of  two generalized Green's operators is again a generalized Green's operator of the product of the corresponding differential operators.
Moreover, if the reverse order law holds, we
can algorithmically determine which boundary problem is solved by the product of two generalized Green's operators.
We present a first implementation of the new algorithms in the framework of integro-differential operators
and illustrate our results by examples, carried out using the \texttt{IntDiffOp}\
package\footnote{Available at \url{http://www.risc.jku.at/people/akorpora/index.html}.}.

Building on results of~\cite{Korporal2012}, we discuss in Section~\ref{sec:Factorization}
a new algorithm and implementation for factoring a generalized boundary problem, such that the
factorization corresponds to the product of the respective generalized Green's operators. We illustrate the algorithm also with examples for differential equations with non-constant coefficients. The right-hand factor computed by this method will always be a regular boundary problem. However,
we also present some first steps for obtaining other possible factorizations in Section~\ref{sec:outlook}.

\section{Symbolic Computation for Boundary Problems} \label{sec:SC}

The algebraic framework for treating linear ordinary boundary problems with symbolic methods is given by the algebra of integro-differential
operators over an ordinary integro-differential algebra. This algebra was introduced
in~\cite{Rosenkranz2005,RosenkranzRegensburger2008a}
as a uniform language to express boundary
problems---meaning differential equations and boundary conditions---as well as their Green's operators,
which are integral operators.
We review the basic properties and refer the reader to~\cite{RosenkranzRegensburger2008a}
and~\cite{RosenkranzRegensburgerTecBuchberger2012} for additional details. See also~\cite{Bavula2011,Bavula2013} for an extensive
study on algebraic properties of integro-differential operators with polynomial coefficients and a single
evaluation (corresponding to initial value problems).

Extending \emph{differential algebras}~\cite{Kolchin1973}, where derivations are linear operators satisfying
the Leibniz rule, \emph{integro-differential algebras} $(\mathcal{F}, \partial, {\textstyle \varint})$ are defined as a differential
algebra $(\mathcal{F}, \partial)$ along with an ``integral'' ${\textstyle \varint}$
that is a linear  right inverse of the derivation $\partial$ and satisfies an algebraic version of
the \emph{integration by parts} formula. For the similar notion of differential Rota-Baxter algebras, see~\cite{GuoKeigher2008} and for a
detailed comparison~\cite{GuoRegensburgerRosenkranz2013}.

We call an integro-differential algebra over a field $K$ \emph{ordinary} if $\ker \partial = K$.
The standard example of an ordinary integro-differential algebra is given by $\mathcal{F} = C^\infty (\mathbb{R})$ with the usual
derivation and the integral operator ${\textstyle \varint} \colon f \mapsto {\textstyle \varint}_a^x f(\xi)\,\mathrm{d} \xi$ for a fixed $a \in \mathbb{R}$.
 We call $\E = 1-{\textstyle \varint} \circ \partial$ the \emph{evaluation} of
$\mathcal{F}$.
For representing not only initial value problems, but arbitrary boundary problems,
we include additional characters (multiplicative linear functionals)
$\E_c\colon f \mapsto f(c)$ at various evaluation
points $c \in \mathbb{R}$.

We write $\galg\langle \partial \rangle$ for the the ring of differential operators with coefficients in $\mathcal{F}$ and $\galg\langle \cum \rangle$ for integral operators
of the form
$\sum_{i=1}^{n} f_i {\textstyle \varint} g_i$ with $f_i, g_i \in \mathcal{F}$.
For a set of characters $\Phi$, the corresponding (two-sided ideal of) \emph{boundary operators} $( \Phi )$ are finite sums
\begin{equation*}
   \sum_{\varphi \in \Phi} \left( \sum_{i \in \mathbb{N}}
      f_{i, \varphi} \, \varphi \partial^i + \sum_{j \in \mathbb{N}} g_{j, \varphi} \varphi {\textstyle \varint} h_{j, \varphi} \right)
\end{equation*}
with $f_{i, \varphi}, g_{j, \varphi}, h_{j, \varphi} \in \mathcal{F}$.
 \emph{Stieltjes boundary conditions} are boundary operators where all the $f_{i, \varphi}\in K$ are constants and $ g_{j, \varphi}=1$, so that they act on $\mathcal{F}$ as linear functionals.

The
integro-differential operators $\galg_{\Phi}\langle \partial, \cum \rangle$ are given as a direct sum of $K$-vector spaces
\begin{equation*}
 \galg_{\Phi}\langle \partial, \cum \rangle = \galg\langle \partial \rangle \dotplus \galg\langle \cum \rangle \dotplus ( \Phi ).
\end{equation*}
The representation of integral operators and boundary conditions is not unique due to linearity of ${\textstyle \varint}$, for
normal forms of integro-differential operators; see~\cite{RosenkranzRegensburger2008a}.

For solving boundary problems, we restrict ourselves to monic (i.e., having leading coefficient $1$) differential
operators  that have a \emph{regular fundamental system} $u_1, \ldots, u_n$, which means that the associated
Wronskian matrix
\begin{equation*}
  W(u_1,\ldots,u_n) = \begin{pmatrix}
    u_1 & \cdots & u_n\\
    u_1' & \cdots & u_n'\\
    \vdots & \ddots & \vdots\\
    u_1^{(n-1)} & \cdots & u_n^{(n-1)}
  \end{pmatrix}
\end{equation*}
is regular. Then the \emph{fundamental right inverse} (solving the initial value problem) can be computed
as an integro-differential operator in $\galg_{\Phi}\langle \partial, \cum \rangle$ by the
\emph{variation of constants} formula
\begin{equation} \label{eq:VarConst}
 \fri{T} = \sum_{i=1}^n u_i {\textstyle \varint} d^{-1}\,d_i,
\end{equation}
 where $d$ is the determinant of the Wronskian matrix, $d_i= \det W_i$,
  and $W_i$ is the matrix obtained from $W$ by replacing the $i$th
  column by the $n$th unit vector.

Algorithms for solving and factoring
regular ordinary boundary problems are described in \cite{RosenkranzRegensburger2008a,RegensburgerRosenkranz2009}.
We recall the basic definitions and results.
A \emph{boundary problem} is defined as a pair $(T, \mathscr{B})$ consisting of
a monic differential operator  $T$ of order $n$ with a regular fundamental system and a space of boundary conditions
$\mathscr{B} = \Ll(\beta_1 , \ldots, \beta_n)$ generated by $n$ linearly independent Stieltjes boundary conditions.
We think of $T$ as a surjective linear operator between suitable $K$-vector spaces of ``functions'' $T\colon V \to W$ and
the boundary conditions $\beta_i$ acting as linear functionals from the dual space $V^*$.
We describe the subspace of functions satisfying the boundary conditions via the orthogonal
\begin{equation*}
 \B^{\perp} = \{f \in V \mid \beta(f) = 0 \text{ for all } \beta \in \mathscr{B} \},
\end{equation*}
i.e., $u \in V$ is a solution of $(T,  \mathscr{B})$ for a given forcing function $f \in W$, if
\begin{equation*}
 Tu=f \quad \text{and} \quad u\in \B^{\perp}.
\end{equation*}
One easily checks that a boundary problem has a unique solution for each forcing function $f$ if and only if
$\Ker T \dotplus \B^{\perp} = V$: The sum $\Ker T + \B^{\perp} =V$ ensures the existence of a solution, the trivial intersection
$\Ker T \cap \B^{\perp} = \{0\}$ implies its uniqueness. Such boundary problems are called \emph{regular}, and
the solution operator $G$ that maps each $f \in W$ to its unique solution $u \in V$ is called \emph{Green's operator}.
The Green's operator is a right inverse of $T$ and can be computed as $G = (1-P)\fri{T}$,
where $P$ is the projector onto $\Ker T$ along $\B^{\perp}$, see also \cite{RosenkranzRegensburger2008a}.
We will use the notation $G = (T, \mathscr{B})^{-1}$ for the Green's operator $G$ of a regular problem $(T, \mathscr{B})$.

Regularity of a boundary problem can be tested via its \emph{evaluation matrix}: Let $(u_1, \ldots, u_n)$
be a basis of $\Ker T$ and $(\beta_1, \ldots, \beta_n)$ a basis of $\mathscr{B}$. Then $(T, \mathscr{B})$ is regular iff
\begin{equation*}
 \beta(u) = \begin{pmatrix}
             \beta_1(u_1) & \ldots & \beta_1(u_n)\\
             \vdots & \ddots & \vdots\\
             \beta_n(u_1) & \ldots & \beta_n(u_n)
            \end{pmatrix} \in K^{n\times n}
\end{equation*}
is regular.

\newpage

For factoring regular boundary problems and their Green's operators, we recall the composition of boundary problems  in~\cite{RegensburgerRosenkranz2009}. This composition corresponds to the product of their Green's operators in reverse order. Since $G_1=(T_1, \mathscr{B}_1)^{-1}$ and $G_2=(T_2, \mathscr{B}_2)^{-1}$ are right inverses of $T_1 \colon V \to W$
and $T_2\colon U \to V$, the product
$G_2G_1$ obviously is a right inverse of $T_1T_2$. Defining the composition of boundary problems as
\begin{equation} \label{eq:compo}
 (T_1, \mathscr{B}_1) \circ (T_2, \mathscr{B}_2) = (T_1T_2, \mathscr{B}_2 + T_2^*(\mathscr{B}_1)),
\end{equation}
where $T_2^* \colon W^* \to V^*$ denotes the transpose map  $\beta \mapsto \beta \circ T_2$, the reverse order~law
\begin{equation} \label{eq:ROL}
 \bigl( (T_1, \mathscr{B}_1) \circ (T_2, \mathscr{B}_2) \bigr)^{-1}= (T_2, \mathscr{B}_2)^{-1} (T_1, \mathscr{B}_1)^{-1}
\end{equation}
always holds for regular problems.
Using this multiplicative structure of boundary problems, it is always possible to split a regular boundary problem $(T, \mathscr{B})$
into regular lower order problems, provided there exists a factorization $T= T_1T_2$ of
the differential operator; see~\cite{RosenkranzRegensburgerTecBuchberger2009}
for a constructive proof that requires only a fundamental system of $T_2$.

We conclude this section with a remark on the ``function spaces'' $V$ and $W$ on which we let the differential
operator $T$ act. The assumption $V=W=\mathcal{F}$ used for example
in~\cite{RosenkranzRegensburger2008a,KorporalRegensburgerRosenkranz2011}---ensuring well-definedness of arbitrary
operations---for some applications is too restrictive. For a given boundary problem $(T, \mathscr{B})$ of order $n$,
it is for example sufficient to consider $n$ times continuously differentiable functions, i.e., $V = C^n[a,b]$
and $W =C[a,b]$, where $a$ and $b$ are the minimal and maximal evaluation point appearing in the Stieltjes conditions
of $\mathscr{B}$.
Similarly, for composing boundary problems $(T_1, \mathscr{B}_1)$ and $(T_2, \mathscr{B}_2)$ of order $m$ and $n$, it suffices to
restrict the domains as to consider $T_2 \colon C^{m+n}[a,b] \to C^m[a,b]$ and $T_1 \colon C^m[a,b] \to C[a,b]$
(with suitable choices of $a$ and $b$).

\section{Generalized Green's Operators} \label{sec:GGOP}

In~\cite{KorporalRegensburgerRosenkranz2011}, several methods of the previous section are generalized
to boundary problems with ``too many'' boundary conditions, meaning that $\ord T < \dim \mathscr{B}$.
These problems are not solvable for all forcing functions $f$; but we keep the condition $\Ker T \cap \B^{\perp} = \{0\}$
to ensure unique solutions. We briefly recall the basic results from~\cite{KorporalRegensburgerRosenkranz2011}.

\begin{definition}
 We call a boundary problem $(T, \mathscr{B})$ \emph{semi-regular} if
 \begin{equation}
  \Ker T \cap \B^{\perp} = \{0\}.
 \end{equation}
 Let $(T, \mathscr{B})$ be a boundary problem with $T\colon V \to W$ and $E \leq W$. We call the triple $(T, \mathscr{B}, E)$ regular, if
 $(T, \mathscr{B})$ is  semi-regular and
 \begin{equation} \label{eq:DirSum2}
  T(\B^{\perp}) \dotplus E = W.
 \end{equation}
 In this case, we call $E$ an \emph{exceptional space} for $(T, \mathscr{B})$ and $(T, \mathscr{B}, E)$ a \emph{generalized
 boundary problem}.
\end{definition}

For simplicity, we also refer to the triple $(T,\mathscr{B},E)$ as a boundary problem and identify
$(T,\mathscr{B})$ with $(T,\mathscr{B},\{0\})$. One easily checks that regularity of $(T, \mathscr{B})$ implies regularity of
$(T, \mathscr{B}, \{0\})$ and vice versa, see also~\cite[Sec.~4.1]{Korporal2012}.

For solving generalized boundary problems in analysis
the forcing function $f$ is projected onto $T(\B^{\perp})$, the space of \emph{admissible forcing functions};  see for example~\cite{Stakgold1979,Loud1970}. This leads to the following algebraic definition of generalized Green's operators.

\begin{definition}
 Let $(T, \mathscr{B}, E)$ be regular with $T \colon V \to W$, and let $Q$ be the projector onto $T(\B^{\perp})$ along $E$.
 Then $u \in V$ is called a \emph{solution} for $f \in W$ if
 \begin{equation} \label{eq:GenSolution}
  Tu =Qf \quad \text{and} \quad u \in \B^{\perp}.
 \end{equation}
 The \emph{generalized Green's operator} maps each $f \in W$ to the unique solution
 according to \eqref{eq:GenSolution}.
\end{definition}

As in the regular case, we use the notation $G = (T, \mathscr{B}, E)^{-1}$ for the generalized Green's operator for a regular boundary problem $(T, \mathscr{B},E)$.

For testing semi-regularity of a boundary problem $(T, \mathscr{B})$ with $\ord T =m$ and $\dim \mathscr{B} =n$,
we have to check whether the associated evaluation matrix
$\beta(u) \in K^{n\times m}$ has full column rank, see also~\cite[Lem.~1]{KorporalRegensburgerRosenkranz2011}.
Condition~\eqref{eq:DirSum2} can be checked analogously, provided an implicit description of $T(\B^{\perp})$ via
\emph{compatibility conditions} $\mathscr{C} = T(\B^{\perp})^{\perp}$:
If $(\gamma_1, \ldots, \gamma_r)$ is a basis of $\mathscr{C}$ and $(e_1, \ldots, e_r)$ is a basis of $E$, then the
corresponding evaluation matrix $\gamma(e) \in K^{r\times r}$ has to be regular.
Moreover, in~\cite[Prop.~1]{KorporalRegensburgerRosenkranz2011} a method is presented to compute a
basis of $\mathscr{C}$; we have
\begin{equation}\label{eq:CompCond}
 \mathscr{C} = G^*(\mathscr{B} \cap (\Ker T)^{\perp})
\end{equation}
for any right inverse $G$ of $T$.
Note that equation~\eqref{eq:CompCond} requires computing the intersection of the finite dimensional space $\mathscr{B}$
with the finite codimensional space $(\Ker T)^{\perp}$, which can be done using the following observation; see for example~\cite{KorporalRegensburger2013}.

\begin{lemma} \label{lem:intersections}
 Let the subspaces $U \leq V$ and $\mathscr{B} \leq V^*$ be generated respectively by $u=(u_1, \ldots, u_m)$
 and $\beta=(\beta_1, \ldots, \beta_n)$.
 Let $k^1, \ldots, k^r \in F^m$ be a basis of $\Ker \beta(u)$, and $\kappa^1, \ldots, \kappa^s \in F^n$ a basis
 of $\Ker (\beta(u))^T$.   Then
 \begin{enumerate}[label=(\roman*)]
\item $U\cap \B^{\perp}$ is generated by
   $\sum_{i=1}^m k^1_i u_i , \ldots, \sum_{i=1}^m k^r_i u_i$ and
 \item $U^{\perp} \cap \mathscr{B} $ is generated by
  $\sum_{i=1}^n \kappa^1_i\beta_i , \ldots, \sum_{i=1}^n \kappa^s_i \beta_i$.
 \end{enumerate}
\end{lemma}

\begin{example} \label{ex:CC}
 We consider the boundary problem \eqref{eq:bpsingular}
 \begin{equation*}
  \bvp{u''(x) = f(x)}{u(1)=u'(1)=u'(0)=0,}
 \end{equation*}
which reads in operator notation
as $(\partial^2, \Ll(\E_1, \E_1\partial, \E_0\partial))$. We employ the
standard integro-differential algebra $C^{\infty}(\mathbb{R})$ with the
  usual derivation and integral operator ${\textstyle \varint} \colon f \mapsto {\textstyle \varint}_0^x f(\xi)\,\mathrm{d}\xi$ that
  is implemented in the \textsc{Maple}\ package \texttt{IntDiffOp}\ as presented
  in~\cite{KorporalRegensburgerRosenkranz2011,KorporalRegensburgerRosenkranz2012}.

 In \cite{KorporalRegensburgerRosenkranz2011}, we have already computed the compatibility conditions and
 the generalized Green's operator with respect to the exceptional space $E=\mathbb{R}$.
 We again carry out the computations in the \texttt{IntDiffOp}\ package, but using
the interface \texttt{IntDiffOperations}\ for input of integro-differential operators. There, we use the symbols $d$ and $a$ for input of the
differential and integral operator, and $e(c)$ for the evaluation at $c \in \mathbb{R}$.
For the \textsc{Maple}\ output, the respective capital letters $\operatorname{D}$, $\operatorname{A}$, and $\E[c]$ are used,
and the non-commutative multiplication of
integro-differential operators is denoted by ``$.$''. The constructors $\operatorname{BP}, \operatorname{GBP}, \operatorname{BC}$ and $\operatorname{ES}$ are used for input
of respectively boundary problems, generalized boundary problems, boundary conditions, and exceptional spaces.
 \begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt}
> with(IntDiffOp):
> with(IntDiffOperations):
> t1 := d^2:    #input differential operator
> b1 := e(1): b2 := e(1).d: b3 := e(0).d:    #boundary conditions
> B1 := BC(b1, b2, b3):
> C1 := CompatibilityConditions(BP(t1, B1)); ~\vspace*{-2ex}
 \[\operatorname{BC}(\E[1].\operatorname{A})\] ~\vspace*{-3ex}
> E1 := ES(1):    #exceptional space
> bp1 := GBP(t1, B1, E1):    #generalized boundary problem
> g1 := GreensOperator(bp1);
\end{alltt}
\begin{equation*}
 x.\operatorname{A} - \operatorname{A}.x + \bigl(-\frac{1}{2}x^2-\frac{1}{2}\bigr).\E_1. \operatorname{A} + \E_1 . \operatorname{A}. x
\end{equation*} ~\vspace*{-2ex}
\end{framed}\normalsize
\end{minipage}
\end{center}

\end{example}

\section{Composition of Generalized Green's Operators} \label{sec:Composition}

In Section~\ref{sec:SC}, we have already recalled the multiplicative structure for boundary problems
introduced
in~\cite{RosenkranzRegensburger2008a,RegensburgerRosenkranz2009}.
In contrast to the regular case, where Green's operators
always satisfy the reverse order law, the situation is more involved for generalized Green's operators
since they are not right inverses of the differential operator.

\begin{proposition}
 \label{prop:GOI}
 Let $(T,\mathscr{B}, E)$ be regular with $T \colon V \to W$ and $G = (T, \mathscr{B}, E)^{-1}$ its generalized Green's operator.
 Then $GTG=G$, that is, $G$ is an outer inverse of $T$.
\end{proposition}

\begin{proof}
  By definition of generalized Green's operators, we  have $Tu=Qf$ for all $f \in W$,
  as well as $Gf=GQf = u$, where $Q$ denotes the projector
 onto $T(\B^{\perp})$ along $E$. Hence $TGf = Tu = Qf$, and $GTGf = GQf = Gf$ for all $f \in W$.
\end{proof}

In terms of generalized inverses, the Green's operator of a regular problem $(T, \mathscr{B}, E)$ can therefore also be defined as the unique outer inverse $G$ of $T$ with $\img G = \B^{\perp}$ and $\Ker G = E$.
In particular, for an outer inverse $G$ of $T$, the boundary problem $(T, (\img G)^{\perp}, \Ker G)$
is regular, and $G$ is its generalized Green's operator; see also \cite[Rmk.~4.7]{Korporal2012}.

The composition $G_2G_1$ of two outer inverses of $T_1$ and $T_2$ is in general not an
outer inverse of the product $T_1T_2$. However, from the above considerations it is clear that if
$G_2G_1$ is an outer inverse of $T_1T_2$,
then computing its kernel and image yields the boundary problem it solves. For a proof of the
following result, see \cite[Thm.~6.2]{KorporalRegensburger2013} or~\cite[Thm.~3.27]{Korporal2012}.

\begin{theorem} \label{prop:ROL}
 Let $(T_1, \mathscr{B}_1, E_1)$ and $(T_2, \mathscr{B}_2, E_2)$ be regular with  $T_1 \colon V \to W$, $T_2 \colon U\to V$
 and $G_1 = (T_1, \mathscr{B}_1, E_1)^{-1}$, $G_2 = (T_2, \mathscr{B}_2, E_2)^{-1}$ their generalized Green's operators.
 If $G_2G_1$ is an outer inverse of $T_1T_2$, the boundary problem
 \begin{equation}\label{eq:Comp1}
  \bigl(T_1T_2, \mathscr{B}_2 + T_2^*(\mathscr{B}_1 \cap E_2^{\perp}), E_1  + T_1(\B^{\perp}_1 \cap E_2) \bigr)
 \end{equation}
 is regular with generalized Green's operator $G_2G_1$.
 Furthermore, the two sums $\mathscr{B}_2 \dotplus T_2^*(\mathscr{B}_1 \cap E_2^{\perp})$ and $E_1  \dotplus T_1(\B^{\perp}_1 \cap E_2)$
 are direct.
\end{theorem}

Based on Equation~\eqref{eq:Comp1}, we define the composition of two arbitrary boundary problems as follows.
\begin{definition}
The \emph{composition} of two boundary problems is defined as
\begin{equation}
\label{eq:compogen}
 (T_1, \mathscr{B}_1, E_1) \circ (T_2,  \mathscr{B}_2, E_2) \\
 = \bigl(T_1T_2, \mathscr{B}_2 + T_2^*(\mathscr{B}_1 \cap E_2^{\perp}), E_1  + T_1(\B^{\perp}_1 \cap E_2) \bigr),
\end{equation}
assuming that the composition $T_1T_2$ is defined.
\end{definition}
This definition clearly reduces to the composition of regular boundary problems~\eqref{eq:compo} when $E_1=E_2=\{0\}$.
The composition of generalized boundary problems is implemented in the following algorithm.

\begin{framed}~\\[-1cm]
\begin{alg}[Composition]~ \label{alg:Comp}
 \begin{description}
 \item[Input] Two  boundary problems $(T_1, \mathscr{B}_1, E_1)$ and $(T_2, \mathscr{B}_2, E_2)$, \\
 $\beta_1, \ldots, \beta_n$ and $\tilde{\beta}_1, \ldots, \tilde{\beta_{\nu}}$ bases of $\mathscr{B}_1$ and $\mathscr{B}_2$, \\
 $e_1, \ldots, e_t$ and $\tilde{e}_1, \ldots,  \tilde{e}_{\tau}$ bases of $E_1$ and $E_2$.
 \item[Output] The composite boundary problem $(T_1, \mathscr{B}_1, E_1)\circ (T_2, \mathscr{B}_2, E_2)$.
\end{description}
\begin{enumerate}
 \item Multiply $T=T_1T_2 \in \mathcal{F}_{\Phi}\langle \partial, {\textstyle \varint} \rangle$.
 \item Compute a basis $\gamma_1, \ldots, \gamma_k$ of $\mathscr{B}_1 \cap E_2^{\perp}$ using Lemma~\ref{lem:intersections}.
 \item Compute a basis $v_1, \ldots, v_{\ell}$ of $\B^{\perp}_1 \cap E_2$ using Lemma~\ref{lem:intersections}.
 \item For $1\leq i \leq k$ multiply $\delta_i = \gamma_i T_2 \in \mathcal{F}_{\Phi}\langle \partial, {\textstyle \varint} \rangle$.
 \item For $1 \leq j \leq \ell$ compute $t_j = T_1(v_j) $.
 \item Compute a basis $\alpha_1, \ldots, \alpha_q$ of $\Ll(\tilde{\beta}_1, \ldots, \tilde{\beta_\nu}, \delta_1, \ldots, \delta_k)$.
 \item Compute a basis $f_1, \ldots, f_r$ of $\Ll(e_1, \ldots, e_t, t_1, \ldots, t_{\ell})$.
 \item Return $(T, (\alpha_1, \ldots, \alpha_q), (f_1, \ldots, f_r))$.
\end{enumerate}
\end{alg}~\\[-1.2cm]
\end{framed}

We are especially interested in the situation of Theorem~\ref{prop:ROL}, that is, the case where the
composition of boundary problems corresponds to the composition of their generalized Green's operators.
For testing when $G_2G_1$ is an outer inverse of $T_1T_2$, we use the following characterization from~\cite{KorporalRegensburger2013,Korporal2012}, which is based on  results from~\cite{GrossTrenkler1998} and \cite{Werner1992}. It gives necessary and sufficient
conditions on the subspaces $\B^{\perp}_1$, $T_2(\B^{\perp}_2)$, $E_2$,  and $T_1^{-1}(E_1)$
such that the revers order law
 \begin{equation}
  \label{eq:ROLgen}
   ((T_1, \mathscr{B}_1, E_1)\circ (T_2, \mathscr{B}_2, E_2))^{-1}=(T_2, \mathscr{B}_2, E_2)^{-1} (T_1, \mathscr{B}_1, E_1)^{-1}
 \end{equation}
for the respective generalized Green's operators holds.
The conditions can be checked using only the input data, in particular, without computing $G_2$ or $G_1$.

\begin{theorem} \label{thm:G2G1OI}
 Let $(T_1, \mathscr{B}_1, E_1)$ and $(T_2, \mathscr{B}_2, E_2)$ be regular with  $T_1 \colon V \to W$, $T_2 \colon U\to V$
 and $G_1 = (T_1, \mathscr{B}_1, E_1)^{-1}$, $G_2 = (T_2, \mathscr{B}_2, E_2)^{-1}$ their generalized Green's operators. Let
 $\mathscr{C}_2 = T_2(\B^{\perp}_2)^{\perp}$ and $K_1 = T_1^{-1}(E_1)$.
The following conditions are equivalent:
\begin{enumerate} [label=(\roman*)]
 \item $G_2G_1$ is an outer inverse of $T_1 T_2$,
 \item $(T_1, \mathscr{B}_1, E_1) \circ (T_2, \mathscr{B}_2, E_2)$ is regular with Green's operator $G_2G_1$,		 	\label{algOI1}
 \item $\mathscr{C}_2 + (\mathscr{B}_1 \cap E_2^{\perp}) \geq \mathscr{B}_1 \cap (E_2 \cap K_1)^{\perp}$, \label{algOI2}
 \item $\mathscr{B}_1 \geq \mathscr{C}_2 \cap (E_2 \cap \B^{\perp}_1)^{\perp} \cap (E_2 \cap  K_1)^{\perp}$, \label{algOI3}
 \item $ K_1  \dotplus (E_2 \cap \B^{\perp}_1) \geq E_2 \cap (\mathscr{B}_1 \cap \mathscr{C}_2)^{\perp}$, \label{algOI4}
 \item $E_2 \geq  K_1 \cap (\mathscr{B}_1 \cap E_2^{\perp})^{\perp} \cap (\mathscr{B}_1 \cap \mathscr{C}_2)^{\perp}$. \label{algOI5}
\end{enumerate}
\end{theorem}

All conditions are formulated so that they can be checked algorithmically with the methods from the end of
Section~\ref{sec:SC}.
For non-algorithmic purposes, also the equivalent dual conditions are useful, as for example
\begin{equation} \label{eq:OI3}
 \B^{\perp}_1 \leq \C^{\perp}_2 \dotplus (E_2 \cap \B^{\perp}_1) \dotplus (E_2 \cap K_1),
 \end{equation}
which is the dual statement of Condition~\ref{algOI3}.

For testing the conditions of Theorem \ref{thm:G2G1OI}, it is necessary to determine
the space $K_1=T_1^{-1}(E_1)$. This can be done
using the identity
\begin{equation} \label{eq:EasyKerComp}
 T^{-1}(E) = G(E) \dotplus \Ker T
 \end{equation}
for the inverse image of a subspace, where $G$ is an arbitrary right inverse  of $T$. It can be verified directly,
see also~\cite[Prop.~A.12]{RegensburgerRosenkranz2009}.  As always, we assume a
regular fundamental system to be given, which allows to construct the fundamental right inverse. Since any right inverse of $T$ is injective and because of the direct sum in \eqref{eq:EasyKerComp}, the output of the following algorithm is indeed a basis of $T^{-1}(E)$.

\begin{framed}~\\[-1cm]
\begin{alg}[Inverse Image]~ \label{alg:preimage}
 \begin{description}
 \item[Input] A monic differential operator $T\colon V \to W$ and a basis $e_1, \ldots, e_r$ of $E \leq W$.
 A regular fundamental system $s_1, \ldots, s_m$ of $T$.
 \item[Output] A basis of $T^{-1}(E)$.
\end{description}
\begin{enumerate}
 \item Compute the fundamental right inverse $\fri{T}$ according to \eqref{eq:VarConst}.
 \item For $1 \leq i \leq r$ compute $k_i = \fri{T} (e_i)$.
 \item Return $s_1, \ldots, s_m, k_1, \ldots, k_r$.
\end{enumerate}
\end{alg}~\\[-1.2cm]
\end{framed}

For testing the necessary and sufficient conditions of Theorem \ref{thm:G2G1OI}, we assume that for
finite-dimensional subspaces of $V$ or $V^*$, we can compute sums and intersections and
check inclusions. For the heuristic approach used in the \texttt{IntDiffOp}\ package, see~\cite[Sec.~7.4]{Korporal2012}.

The following algorithm implements a test of Condition~\ref{algOI2}, the others can be
implemented similarly. (None of them seems to be particularly preferable from computational aspects.)

\begin{framed}~\\[-1cm]
\begin{alg}[Check Reverse Order Law]~ \label{alg:checkROL}
 \begin{description}
 \item[Input] Two  boundary problems $(T_1, \mathscr{B}_1, E_1)$ and $(T_2, \mathscr{B}_2, E_2)$, \\
 $\beta_1, \ldots, \beta_n$ and $\tilde{\beta}_1, \ldots, \tilde{\beta_{\nu}}$ bases of $\mathscr{B}_1$ and $\mathscr{B}_2$, \\
 $e_1, \ldots, e_t$ and $\tilde{e}_1, \ldots,  \tilde{e}_{\tau}$ bases of $E_1$ and $E_2$. \\
 Regular fundamental systems $s_1, \ldots, s_m$ of $T_1$ and $\tilde{s}_1, \ldots, \tilde{s}_{\ell}$ of $T_2$.
 \item[Output] \textbf{true} if the reverse order law holds and \textbf{false} otherwise.
\end{description}
\begin{enumerate}
 \item Compute a basis of $T_1^{-1}(E_1)$ with Algorithm~\ref{alg:preimage}.
 \item Compute a basis of $J = E_2 \cap T_1^{-1}(E_1)$.
 \item Compute a basis of $B = \mathscr{B}_1 \cap J^{\perp}$ using Lemma~\ref{lem:intersections}.
 \item Compute a basis of $K = \mathscr{B}_1 \cap E_2^{\perp}$ using Lemma~\ref{lem:intersections}.
 \item Compute the compatibility conditions $\gamma_1, \ldots, \gamma_r$ of $(T_2, \mathscr{B}_2)$ due to \eqref{eq:CompCond}.
 \item Compute $C = \Ll(\gamma_1, \ldots, \gamma_r) + K$.
 \item If $B\leq C$ return \textbf{true}, else return \textbf{false}.
\end{enumerate}
\end{alg}~\\[-1.2cm]
\end{framed}

\begin{example} \label{ex:GIpaper}
As a first example, we consider the generalized boundary problem
 $(\operatorname{D}^2, \Ll(\E_1, \E_1\operatorname{D}, \E_0\operatorname{D}), \mathbb{R})$
 from Example \ref{ex:CC}. As a second boundary problem, we consider the differential operator
$T_2 = \operatorname{D}^2 - 1$, also with the boundary conditions $\Ll(\E_1, \E_1\operatorname{D}, \E_0\operatorname{D})$, or, in usual notation
\begin{equation*}
 \bvp{u''(x) - u(x) = f(x)}{u(1)=u'(1)=u'(0)=0.}
\end{equation*}
As we will see below, the forcing function $f$ has to satisfy the  compatibility condition
\begin{equation*}
 \textstyle{\int_0^1} \exp(-\xi)f(\xi)\,\mathrm{d}\xi + \textstyle{\int_0^1} \exp(\xi)f(\xi)\,\mathrm{d}\xi =0,
\end{equation*}
 so that $E_2= \Ll(x)$ is
a suitable exceptional space for the second boundary problem.
Moreover, we compute the inverse image $K_1=T_1^{-1}(E_1)$ for testing the
conditions of Theorem~\ref{thm:G2G1OI}.
\begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt}
> t2 := d^2-1: B2 := B1:
> C2 := CompatibilityConditions(BP(t2, B2))); ~\vspace*{-2ex}
 \[ \operatorname{BC}(\E[1].\operatorname{A}.\exp(-x)+\E[1].\operatorname{A}.\exp(x))  \] ~\vspace*{-3ex}
> E2 := ES(x): bp2 := GBP(t2, B2, E2):
> IsRegular(GBP(t2, B2, E2)); ~\vspace*{-3ex}
\[true\] ~\vspace*{-3ex}
> K1 := InverseImage(t1, E1);
\end{alltt} ~\vspace*{-3ex}
\begin{equation*}
 \operatorname{ES}(1, x, x^2)
\end{equation*} ~\vspace*{-3ex}
\end{framed}\normalsize
\end{minipage}
\end{center}

In this case $E_2 =\Ll(x) \leq \Ll(1,x,x^2) =K_1$, so Condition (iv) of Theorem~\ref{thm:G2G1OI}
is trivially satisfied. We check that the reverse order law holds for the respective Green's operators.

\begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt} ~\vspace*{-2ex}
 > p1 := MultiplyBoundaryProblem(bp1, bp2);
\end{alltt} \scriptsize
\begin{equation*}
 \operatorname{GBP}(\operatorname{D}^4-\operatorname{D}^2, \operatorname{BC}(\E[0].\operatorname{D}, \E[0].\operatorname{D}^3-\E[1].\operatorname{D}^3, \E[1], \E[1].\operatorname{D}, \E[1].\operatorname{D}^2-\E[1].\operatorname{D}^3), \operatorname{ES}(1))
\end{equation*} \small
\begin{alltt}
 > g := GreensOperator(p1):
 > g2 := GreensOperator(bp2):
 > IsZero(g-g2.g1);    #check reverse order law ~\vspace*{-3ex}
 \[true\] ~\vspace*{-3ex}
 \end{alltt} ~\vspace*{-3ex} ~\vspace*{-3ex} ~\vspace*{-3ex}
\end{framed}\normalsize
\end{minipage}
\end{center}

Since for the boundary problems in the previous example, we have $T_1T_2 = T_2T_1 = \operatorname{D}^4 -\operatorname{D}^2$, we can also
consider the product of Green's operators in reverse order, that is, test if $(T_1, \mathscr{B}_1, E_1)^{-1}(T_2, \mathscr{B}_2, E_2)^{-1}$
is an outer inverse of $T_1T_2$.

\begin{example} \label{ex:GI2}
We follow the steps of Algorithm \ref{alg:checkROL}, interchanging the indices accordingly.
Recall from Example~\ref{ex:CC} that $\mathscr{C}_1$ is generated by $\E_1\operatorname{A}$ and
that we have chosen the exceptional space $E_1 = \mathbb{R}$.

 \begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt}
 > K2 := InverseImage(t2, E2);
\end{alltt}
\begin{equation*}
 \operatorname{ES}(x, \exp(x), \exp(-x))
\end{equation*}
\begin{alltt}
 > J := Intersection(E1, K2); ~\vspace*{-2ex}
 \[ \operatorname{ES}() \] ~\vspace*{-2ex}
 > B := Intersection(B2, J, space = dual);
 \[ \operatorname{BC}(\E[1], \E[1].\operatorname{D}, \E[0].\operatorname{D})  \]~\vspace*{-2ex}
 > K := Intersection(B2, E1, space = dual);
 \[ \operatorname{BC}(\E[1].\operatorname{D}, \E[0].\operatorname{D})  \]~\vspace*{-2ex}
 > TestComposition(bp2, bp1);    #check reverse order law ~\vspace*{-2ex}
 \[false\]~\vspace*{-3ex} ~\vspace*{-3ex}
\end{alltt}
\end{framed}\normalsize
\end{minipage}
\end{center}
\end{example}
We see that in the above example Algorithm \ref{alg:checkROL} returns \emph{false} since
the inclusion
\begin{equation*}
 \mathscr{C}_1 + (\mathscr{B}_2 \cap E_1^{\perp}) \geq \mathscr{B}_2 \cap (E_1 \cap K_2)^{\perp}
\end{equation*}
from Theorem \ref{thm:G2G1OI} (ii) is not satisfied: the left-hand side does not contain
the evaluation $\E_1$.
We also verify directly that the product $G_1G_2$ indeed is not an outer inverse of $T_1T_2$.

\begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt} ~\vspace*{-2ex}
 > t := t2.t1: g:= g1.g2:
 > IsZero((g.t).g-g); ~\vspace*{-3ex}
  \[false\] ~\vspace*{-3ex}
 > p2 := MultiplyBoundaryProblem(bp2, bp1);
\end{alltt} \scriptsize
\begin{equation*}
 \operatorname{GBP}(\operatorname{D}^4-\operatorname{D}^2, \operatorname{BC}(\E[0].\operatorname{D}, \E[0].\operatorname{D}^3, \E[1], \E[1].\operatorname{D}, \E[1].\operatorname{D}^3), \operatorname{ES}(x))
\end{equation*} \small
 \begin{alltt}
 > IsRegular(p2); ~\vspace*{-3ex}
 \[true\]~\vspace*{-3ex} ~\vspace*{-3ex}
\end{alltt}
\end{framed}\normalsize
\end{minipage}
\end{center}

Although $G_1G_2$ is not an outer inverse of $T_2T_1$, in this case the composite boundary problem
$(T_2, \mathscr{B}_2, E_2) \circ (T_1, \mathscr{B}_1, E_1)$ is regular. In general, we do not even obtain semi-regularity
of $(T_1T_2, \mathscr{B}_2 + T_2^*(\mathscr{B}_1 \cap E_2^{\perp}))$, see~\cite[Thm.~4.21]{Korporal2012} for more details.
\end{example}

\section{Factorization of Generalized Boundary Problems} \label{sec:Factorization}

In this section, we discuss certain classes of factorizations of generalized boundary problems.
We start with a regular boundary problem $(T, \mathscr{B}, E)$ and a factorization $T= T_1T_2$ of the underlying differential operator.
The factorization algorithm presented in this section works for arbitrary factorizations of $T$, regardless of the
coefficient domain, as long as a regular fundamental system of $T_2$ is known.
In our package, we use the function \texttt{DFactor} from the \textsc{Maple}\ \texttt{DETools} package to factor differential
 operators with rational coefficients (Example \ref{ex:RatCoeff}), unless a particular factorization is specified in the input.

As in \cite{RegensburgerRosenkranz2009} for regular boundary problems, the overall goal would be to characterize all regular boundary problems $(T_1, \mathscr{B}_1, E_1)$ and $(T_2, \mathscr{B}_2, E_2)$ such that
\begin{equation*}
 (T, \mathscr{B}, E) = (T_1, \mathscr{B}_1, E_1) \circ (T_2, \mathscr{B}_2, E_2).
\end{equation*}
For generalized boundary problems, we additionally have to require that the reverse order law
\begin{equation*}
 ((T_1, \mathscr{B}_1, E_1) \circ (T_2, \mathscr{B}_2, E_2))^{-1} = (T_2, \mathscr{B}_2, E_2)^{-1} (T_1, \mathscr{B}_1, E_1)^{-1}
\end{equation*}
holds, which is always valid for regular boundary problems~\eqref{eq:ROL}. Due to the structure of  the composite~\eqref{eq:compogen}, where information about $\mathscr{B}_1$ gets lost when intersecting
with $E_2$, and the rather involved interactions of the subspaces $\B^{\perp}_1, K_1, \C^{\perp}_2$ and $E_2$ in Theorem~\ref{thm:G2G1OI}, it is a difficult task to describe \emph{all} possible factorizations.

In the following, we discuss a special case of factoring a regular boundary problem $(T, \mathscr{B}, E)$:
As a right factor, we strive for a regular problem $(T_2, \mathscr{B}_2)$, meaning that $E_2=\{0\}$. Then
the reverse order law is trivially satisfied,
and the composite takes the easier form
\begin{equation*}
 (T_1, \mathscr{B}_1, E_1) \circ (T_2, \mathscr{B}_2) = (T_1T_2, \mathscr{B}_2 \dotplus T_2^*(\mathscr{B}_1), E_1).
\end{equation*}
The existence of such a factorization was proven in~\cite{Korporal2012}. It relies on the fact that
for a semi-regular boundary problem $(T, \mathscr{B})$ and a factorization $T=T_1T_2$, there always exists
$\mathscr{B}_2 \leq \mathscr{B}$ such that $(T_2, \mathscr{B})$ is regular.
(For ordinary differential equations, this can be seen immediately by inspecting the evaluation matrix.)

\begin{theorem} \label{thm:FactReg}
 Let $(T, \mathscr{B})$ be a semi-regular boundary problem and let $T= T_1T_2$ be a factorization of $T$ into
 monic differential operators.
   For each exceptional space $E$ of $(T, \mathscr{B})$ there exists a unique regular boundary problem
 $(T_1, \mathscr{B}_1, E)$ such that for each $\mathscr{B}_2 \leq \mathscr{B}$ for which $(T_2, \mathscr{B}_2)$ is regular, we have
 \begin{equation} \label{eq:RegLeft}
  (T, \mathscr{B}, E) = (T_1, \mathscr{B}_1, E) \circ (T_2, \mathscr{B}_2) = (T_1T_2, \mathscr{B}_2 \dotplus T_2^*(\mathscr{B}_1), E)
 \end{equation}
 and $(T, \mathscr{B}, E)^{-1} =  (T_2, \mathscr{B}_2)^{-1} (T_1, \mathscr{B}_1, E)^{-1}$. The boundary conditions of the left
 factor are given by $\mathscr{B}_1 = H_2^*(\mathscr{B} \cap (\Ker T_2)^{\perp})$, where $H_2$ is
 an arbitrary right inverse of $T_2$.

\end{theorem}

\begin{example} \label{ex:fac1}
We consider the factorization of the boundary problem \texttt{p2} from Example~\ref{ex:GI2} with
$\operatorname{D}^4-\operatorname{D}^2 = (\operatorname{D}^2 -1)\cdot \operatorname{D}^2$:
\begin{multline*}
 (\operatorname{D}^4-\operatorname{D}^2, \Ll(\E_0\operatorname{D}, \E_0\operatorname{D}^3, \E_1, \E_1\operatorname{D}, \E_1\operatorname{D}^3), \Ll(x)) \\
 = (\operatorname{D}^2 -1, \Ll(\E_0\operatorname{D}, \E_1 \operatorname{D}, \E_1\operatorname{A}), \Ll(x)) \circ
 (\operatorname{D}^2, \Ll(\E_0\operatorname{D}, \E_1)).
\end{multline*}
The following \textsc{Maple}\ session shows that the reverse order law holds for the respective Green's operators.

\begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt} ~\vspace*{-2ex}
 > p2; ~\vspace*{-2ex}
\end{alltt}  \scriptsize
\begin{equation*}
 \operatorname{GBP}(\operatorname{D}^4-\operatorname{D}^2, \operatorname{BC}(\E[0].\operatorname{D}, \E[0].\operatorname{D}^3, \E[1], \E[1].\operatorname{D}, \E[1].\operatorname{D}^3), \operatorname{ES}(x))
\end{equation*} \small
\begin{alltt}
 > g := GreensOperator(p2):
 > f1, f2 := FactorBoundaryProblem(p2, t2, t1); ~\vspace*{-3ex}
 \end{alltt}\scriptsize
 \begin{equation*}
  \operatorname{GBP}(D^2-1, \operatorname{BC}(\E[0].\operatorname{D}, \E[1].\operatorname{D}, \E[1].\operatorname{A}), \operatorname{ES}(x)), \operatorname{BP}(\operatorname{D}^2, \operatorname{BC}(\E[0].\operatorname{D}, \E[1]))
 \end{equation*}\small
\begin{alltt}
 > g1 := GreensOperator(f1): g2:= GreensOperator(f2):
 > IsZero(g-g2.g1); ~\vspace*{-3ex}
 \[true\]
\end{alltt} ~\vspace*{-3ex} ~\vspace*{-3ex} ~\vspace*{-3ex}
\end{framed}\normalsize
\end{minipage}
\end{center}
\end{example}

We now show how to construct the above factorization algorithmically,
generalizing the method presented in  \cite{RosenkranzRegensburgerTecBuchberger2009}.
We assume a given factorization $T= T_1T_2$ of the differential operator and a regular fundamental system of
$T_2$, which is needed in Step 4 to construct a right inverse of $T_2$.

\begin{framed}~\\[-1cm]
\begin{alg}[Right Regular Factorization]~ \label{alg:RightFact}
\begin{description}
 \item[Input] A regular boundary problem $(T, \mathscr{B}, E)$, \\
 $\beta_1, \ldots, \beta_{n}$ a basis of $\mathscr{B}$,
 $e_1, \ldots, e_r$ a basis of $E$.  \\ A factorization $T = T_1T_2$,  a reg.~fundamental system $s_1, \ldots, s_{\mu}$
 of $T_2$.
 \item[Output] Two regular boundary problems $(T_1, \mathscr{B}_1, E)$ and $(T_2, \mathscr{B}_2)$ with $(T, \mathscr{B}, E) = (T_1, \mathscr{B}_1, E) \circ (T_2, \mathscr{B}_2)$.
\end{description}
\begin{enumerate}
 \item Compute the evaluation matrix $M = \beta(s) \in F^{n \times \mu}$.
 \item Compute $S = (s_{i,j}) \in F^{n \times n}$ s.t. ~$SM$ is in reduced row echelon form. \label{step:LUDec}
 \item For $1 \leq i \leq n$ set $\tilde{\beta}_i = \sum_{k=1}^n s_{i,k} \beta_k$.
 \item Compute a right inverse $H_2$ of $T_2$ according to \eqref{eq:VarConst}.
 \item For $\mu+1 \leq j \leq n$ multiply $\alpha_{j-\mu} = \tilde{\beta}_j H_2 \in \mathcal{F}_{\Phi}\langle \partial, {\textstyle \varint} \rangle$. \label{step:Hright}
 \item Return $(T_1, (\alpha_1, \ldots, \alpha_{n-\mu}), (e_1, \ldots, e_r))$ and
 $(T_2, (\tilde{\beta}_1, \ldots, \tilde{\beta}_\mu))$.
\end{enumerate}
\end{alg}~\\[-1.2cm]
\end{framed}

\begin{theorem}
 Let $(T, \mathscr{B}, E)$ be regular and $T = T_1T_2$.
 The boundary problems $(T_1, \mathscr{B}_1, E)$ and $(T_2, \mathscr{B}_2)$ computed by Algorithm~\ref{alg:RightFact} are
 regular and satisfy
 \begin{equation*}
  (T, \mathscr{B}, E) = (T_1, \mathscr{B}_1, E) \circ (T_2, \mathscr{B}_2).
 \end{equation*}
\end{theorem}

\begin{proof}
 In view of Theorem~\ref{thm:FactReg}, we only have to show that $(T_2, \Ll(\tilde{\beta}_1, \ldots , \tilde{\beta}_{\mu}))$
 is regular with $ \tilde{\beta}_i \in \mathscr{B}$ for $1\leq i \leq \mu$,  and that the Stieltjes conditions
 $\tilde{\beta}_{\mu+1}, \ldots, \tilde{\beta}_n$ computed in Step 3, are a basis of $\mathscr{B} \cap (\Ker T_2)^{\perp}$.

 First we observe that in Step 3 we obviously have $\tilde{\beta}_i \in \mathscr{B}$ for all $i$. Since $\Ker T_2 \leq \Ker T$ and since $(T, \mathscr{B})$ is semi-regular, $(T_2, \mathscr{B})$ is also semi-regular.
 Hence the evaluation matrix $M$ computed in Step 1 has rank $\mu$,
 and the first $\mu$ rows of $SM$ in Step 2 give the $\mu \times \mu$ identity matrix.
 Since $SM$ is the evaluation matrix $\tilde{\beta}(s)$---meaning that the $\tilde{\beta}$ are applied to the fundamental system of
 $\Ker T_2$---the boundary problem $(T_2, \Ll(\tilde{\beta}_1, \ldots , \tilde{\beta}_{\mu}))$ is regular.

 The lower part of $SM$ is the $(n-\mu) \times \mu$ zero matrix, hence we know that
 \begin{equation*}
  \Ll(\tilde{\beta}_{\mu+1}, \ldots, \tilde{\beta}_n) \leq \mathscr{B} \cap (\Ker T_2)^{\perp}.
 \end{equation*}
 Moreover, $\tilde{\beta}_{\mu+1}, \ldots, \tilde{\beta}_n$ are linearly independent, since $S$ is regular.
 For proving that the spaces are equal, it suffices to show that $\dim (\mathscr{B} \cap (\Ker T_2)^{\perp})= n - \mu$.
 Since $\mathscr{B}$ is finite dimensional, also $\dim (\mathscr{B} \cap (\Ker T_2)^{\perp}) < \infty$, and we have
 \begin{equation*}
  \dim (\mathscr{B} \cap (\Ker T_2)^{\perp}) = \codim(\mathscr{B} \cap (\Ker T_2)^{\perp})^{\perp}
  = \codim (\B^{\perp} + \Ker T_2),
 \end{equation*}
 where we use the duality principle for switching between a vector space and its dual
 as explained in \cite[Sec.~A.1]{RegensburgerRosenkranz2009}. Furthermore, \cite[Cor.~A.15]{RegensburgerRosenkranz2009}
 yields
 \begin{equation*}
  \codim(\mathscr{B} \cap (\Ker T_2)^{\perp})^{\perp} = \dim (\Ker T_2 \cap \B^{\perp} ) + \codim \B^{\perp} - \dim \Ker T_2.
 \end{equation*}
 Since $(T_2, \mathscr{B})$ is semi-regular, the first summand vanishes, and moreover using that $\codim \B^{\perp} = \dim \mathscr{B}$, we
 obtain $ \dim (\mathscr{B} \cap (\Ker T_2)^{\perp}) = n- \mu$.
\end{proof}

\begin{example} \label{ex:fac2}
 We demonstrate the steps of the algorithm in detail, continuing Example \ref{ex:fac1}.
 Obviously, a regular fundamental system of $\operatorname{D}^2$ is given by $(1, x)$ and a right inverse of $\operatorname{D}^2$
 is $\operatorname{A}^2$. The evaluation matrix
 $M$ and the transformation matrix $S$ are given as
 \begin{equation*}
  M= \begin{pmatrix}
   0 & 1 \\ 0 &0\\1 &1\\0 &1\\0&0
  \end{pmatrix} \quad \text{and} \quad
 S= \begin{pmatrix}
   -1 &0&1&0&0\\1&0&0&0&0\\0&1&0&0&0\\-1&0&0&1&0\\0&0&0&0&1
  \end{pmatrix}.
 \end{equation*}
 Hence we compute the $\tilde{\beta}$ as follows
 \begin{equation*}
  \tilde{\beta}_1 = \E_1 - \E_0\operatorname{D}, \;
  \tilde{\beta}_2 = \E_0\operatorname{D},\;
  \tilde{\beta}_3 = \E_0 \operatorname{D}^3,\;
  \tilde{\beta}_4 = \E_1\operatorname{D} - \E_0\operatorname{D},\;
  \tilde{\beta}_5 = \E_1\operatorname{D}^3.
 \end{equation*}
 The space of boundary conditions for the right-hand factor is spanned by $\tilde{\beta}_1$ and $\tilde{\beta}_2$, which means
 that $\mathscr{B}_2 = \Ll(\E_1, \; \E_0\operatorname{D})$. We multiply $\tilde{\beta}_3, \tilde{\beta}_4$ and $\tilde{\beta}_5$ by $\operatorname{A}^2$ to
 obtain a basis of $\mathscr{B}_1$. Since $\operatorname{D}\operatorname{A}=1$ and $\E_0\operatorname{A}=0$, this yields
 $\mathscr{B}_1 = \Ll(\E_0\operatorname{D}, \E_1\operatorname{A}, \E_1\operatorname{D})$, as already seen in Example \ref{ex:fac1}.
 \end{example}

 The previous example shows that the computation of the boundary conditions $\mathscr{B}_1$ and $\mathscr{B}_2$ reduces to a linear algebra
 problem, which in our implementation is solved by the \textsc{Maple}\ procedures from the \texttt{LinearAlgebra}
 package. In the \texttt{IntDiffOp}\ package, the most time-consuming operations---factoring the differential operator and computing
 a fundamental system for the right factor---are done in preprocessing steps via the \textsc{Maple}\ commands \texttt{dsolve} and
 \texttt{DFactor}.

 \begin{example}\label{ex:RatCoeff}

  On the interval $[0, 1]$, we consider the differential operator
  \begin{multline*}
   \operatorname{D}^4 + \frac{5 x^2+4 x + 1}{(x+1)(x^2+1)} \operatorname{D}^3 + \frac{x^7+x^6+2 x^5+2 x^4 -x^3-5x^2+14x+10}{(x+1)(x^2+1)^2} \operatorname{D}^2\\
   + \frac{2(2x^8+2x^7+4x^6+4x^5 + x^4+2x^3-14x^2-16x+3)}{(x^2+1)^3(x+1)} \operatorname{D} \\
   + \frac{2(x^7+x^6+2x^5+2x^4 +5x^3+7x^2-4x-2)}{(x^2+1)^3(x+1)}
  \end{multline*}
  with coefficients in $\mathbb{Q}(x)$.

  \begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt} ~\vspace*{-2ex} \scriptsize
 > a3 := (5*x^2+4*x+1)/((x+1)*(x^2+1)):
 > a2 := (x^7 + x^6 + 2*x^5 + 2*x^4-x^3-5*x^2+14*x+10)/((x+1)*(x^2+1)^2):
 > a1 := 2*(2*x^8+2*x^7+4*x^6+4*x^5+x^4+2*x^3-14*x^2-16*x+3)/((x^2+1)^3*(x+1)):
 > a0 := 2*(x^7+x^6+2*x^5+2*x^4+5*x^3+7*x^2-4*x-2)/((x^2+1)^3*(x+1)):
 > t := d^4 + a3.(d^3) + a2.(d^2) + a1.d + a0:
 > b1 := e(0): b2 := e(0).d: b3 :=e(0).(d^2): b4 := e(1): b5 := e(1).d: \small
 > bp := GBP(t, BC(b1, b2, b3, b4, b5), ES(1)):
 > FactorBoundaryProblem(bp);
\end{alltt}  \scriptsize
\begin{align*}
&\operatorname{GBP}(x^2+\frac{1}{1+x} . \operatorname{D}+\operatorname{D}^2, \operatorname{BC}(\E[0], \E[1] . \operatorname{A} . x^2+\E[1] . \operatorname{A}, \E[1] . \operatorname{A} . x^3+\E[1] . \operatorname{A} . x), \operatorname{ES}(1)),\\
&\operatorname{BP}(\frac{2x}{x^2+1}+\operatorname{D}, \operatorname{BC}(\E[0])),\;\;\operatorname{BP}(\frac{2x}{x^2+1}+\operatorname{D}, \operatorname{BC}(\E[0]))
\end{align*}
\end{framed}\normalsize
\end{minipage}
\end{center}

 \end{example}

 We conclude this section with an example for a more general differential operator, which cannot be factored
 with the \texttt{DFactor} command. In this case, a factorization of the differential operator has to be provided
 in the input data.

 \begin{example} \label{ex:exp}
 We consider a differential operator on $\Omega=(0, \infty)$ with coefficients in $C^{\infty}(\Omega)$.
 \begin{equation*}
  \operatorname{D}^2 - \frac{\exp(x) + \exp(2x) -1}{\exp(x)-1} \operatorname{D} + \frac{\exp(2x)}{\exp(x)-1}
 \end{equation*}
with boundary conditions $\E_1, \E_2, \E_3$, and the factorization
\begin{equation*}
 T_1 = \operatorname{D}- \frac{\exp(2x)}{\exp(x)-1}\quad \text{and} \quad T_2 = \operatorname{D}-1.
\end{equation*}

   \begin{center}
\begin{minipage}{0.999\textwidth}
\begin{framed}~\\[-1cm]\small
\begin{alltt}
 > t := d^2-(exp(x)+exp(2*x)-1)/(exp(x)-1).d+exp(2*x)/(exp(x)-1):
 > t1 := d-exp(2*x)/(exp(x)-1):
 > t2 := d-1:
 > b1 := e(1): b2 := e(2): b3 := e(3):
 > bp := GBP(t, BC(b1, b2, b3), ES(1)):
 > FactorBoundaryProblem(bp, t1, t2);
\end{alltt}  \scriptsize
\begin{align*}
&\operatorname{GBP}(\frac{-\exp(2x)}{\exp(x)-1} +\operatorname{D}, \operatorname{BC}(\E[1].\operatorname{A}.\exp(-x), \E[2].\operatorname{A}.\exp(-x) -\E[3].\operatorname{A}.\exp(-x) ), \operatorname{ES}(1)),\\
&\operatorname{BP}(-1+\operatorname{D}, \operatorname{BC}(\E[1]))
\end{align*}
\end{framed}\normalsize
\end{minipage}
\end{center}

 \end{example}

\section{Conclusion and Outlook}
\label{sec:outlook}

As outlined in the previous section, describing all possible factorizations of a generalized boundary problem along a fixed factorization of the differential operator is quite involved.
 Nevertheless, it would sometimes be preferable to have the generalized factor on the right,
for which we assume symbolic solutions for the differential operator.
In this section, we describe some first steps in this direction that rely on the following result.

 \begin{theorem} \label{thm:ChoseE2}
Let $(T_1, \mathscr{B}_1)$ and $(T_2, \mathscr{B}_2)$ be semi-regular with $T_1 \colon V \to W$ and $T_2 \colon U \to V$.
Then there exists an exceptional space $E_2 \leq V$ of $(T_2, \mathscr{B}_2)$ such that the
respective generalized Green's operators satisfy the reverse order law for all possible
exceptional spaces $E_1 \leq W$ of $(T_1, \mathscr{B}_1)$.
\end{theorem}

\begin{proof}
 Let $V_1$ be a complement of $\B^{\perp}_1 \cap T_2(\B^{\perp}_2)$ in $\B^{\perp}_1$, i.e.,
 \[\B^{\perp}_1 = \bigl(\B^{\perp}_1 \cap T_2(\B^{\perp}_2) \bigr) \dotplus V_1.\]
 Since $V_1 \leq \B^{\perp}_1$ and because of the direct sum, we then have
 \[V_1 \cap T_2(\B^{\perp}_2)
  = \bigl( V_1 \cap T_2(\B^{\perp}_2) \bigr) \cap \B^{\perp}_1
  = V_1 \cap \bigl( T_2(\B^{\perp}_2)  \cap \B^{\perp}_1 \bigr) = \{0\}.
 \]
Enlarging $V_1$ to a complement $E_2$ of $T_2(\B^{\perp}_2)$, so that $V_1 \leq E_2$ and
$T_2(\B^{\perp}_2) \dotplus E_2 = V$ yields
\[ \B^{\perp}_1 = \bigl(\B^{\perp}_1 \cap T_2(\B^{\perp}_2) \bigr) \dotplus V_1 \leq T_2(\B^{\perp}_2) + (E_2 \cap \B^{\perp}_1).\]
Hence Condition~\eqref{eq:OI3} is satisfied.
\end{proof}

The previous proof is not constructive; choosing exceptional spaces for arbitrary Stieltjes
boundary conditions leads to an interpolation problem with integral conditions. However,  one can apply the
following strategy to obtain a factorization of a semi-regular boundary problem $(T, \mathscr{B})$ into
a regular problem $(T_1, \mathscr{B}_1)$ as a left factor
and a generalized boundary problem as a right factor. In this case we will ignore the exceptional
space in the beginning and (try to) choose it accordingly in the end.

\begin{enumerate}
 \item Apply (a version of) Algorithm~\ref{alg:RightFact} to compute a semi-regular factor $(T_1, \mathscr{B}_1)$ and
  a regular factor $(T_2, \mathscr{B}_2)$.
 \item Let $\alpha_1, \ldots, \alpha_n$ be a basis of $\mathscr{B}_1$. Choose a subset (wlog we write $\alpha_{1}, \ldots, \alpha_{m}$),
 such that $(T_1, \Ll(\alpha_{1}, \ldots, \alpha_{m}))$ is regular. Such a subset exists by Lemma~\ref{lem:intersections}
 and can also be computed.
 \item Compute $T_2^*(\alpha_{m+1}), \ldots, T_2^*(\alpha_{n})$.
 \item Since $(T_2, \mathscr{B}_2)$ is regular, $(T_2, \mathscr{B}_2 + \Ll(T_2^*(\alpha_{m+1}), \ldots, T_2^*(\alpha_{n}))$ is
 semi-regular, and from Theorem \ref{thm:ChoseE2} we obtain an appropriate exceptional space.
\end{enumerate}

\begin{example}
 In Example \ref{ex:fac2}, we have computed the spaces of boundary conditions
 \begin{equation*}
  \mathscr{B}_1 = \Ll(\E_0\operatorname{D}, \E_1 \operatorname{D}, \E_1\operatorname{A}) \quad \text{and} \quad
  \mathscr{B}_2 = \Ll(\E_0\operatorname{D}, \E_1)
 \end{equation*}
 for the differential operators $T_1 = \operatorname{D}^2 -1$ and $T_2 = \operatorname{D}^2$. The boundary problem $(T_1, \Ll\E_0\operatorname{D}, \E_1\operatorname{D})$ is regular,
 and we compute
 \begin{equation*}
   T_2^*(\E_1\operatorname{A})= \E_1\operatorname{A}\operatorname{D}^2 = \E_1\operatorname{D} - \E_0\operatorname{D}.
 \end{equation*}
 Computation of the compatibility conditions for the modified right-hand factor
 $(T_2, \Ll(\E_0\operatorname{D}, \E_1, \E_1\operatorname{D}))$ yields $\mathscr{C}_2 = \Ll(\E_1\operatorname{A})$, hence we may choose for example
 $E_2 = \Ll(\exp(x))$. Then, since $E_2 \leq \Ker T_1$, Condition~\ref{algOI2} of Theorem~\ref{thm:G2G1OI}
 is trivially fulfilled.
\end{example}

Closer investigation of Algorithm \ref{alg:RightFact} and the above procedure indicates some redundancy in first applying
$H_2^*$ and afterwards $T_2^*$. However, we cannot decide in advance for which choices of $\alpha_i$ the boundary problem
$(T_1, \Ll(\alpha_1, \ldots, \alpha_m))$ will be regular. We will study this and related question for obtaining different factorizations of generalized boundary problems in future work.

We are also investigating algebraic and algorithmic aspects of generalized Green's  operators for certain classes of linear partial differential equations and boundary conditions. While the linear algebra setting used in this paper is in principle applicable to partial differential equations, examples and constructive methods for singular partial boundary problems are to be studied. In this context, we refer to~\cite{RegensburgerRosenkranz2009,RosenkranzRegensburgerTecBuchberger2009,Tec2011,RosenkranzPhisanbut2013}
for regular partial boundary problems. The setting in~\cite{RosenkranzPhisanbut2013} includes also  inhomogeneous boundary conditions and a rewrite system for partial integro-differential operators (PIDOS) including linear substitution.

\bigskip

\noindent \textbf{Acknowledgements.} We would like to thank the anonymous referees for their numerous helpful comments. G.R.~was partially supported by the Austrian Science Fund (FWF): J 3030-N18.

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Thousands of devotees from the Midwestern states, especially from Chicago land used to travel to Pittsburgh to have darshan of Sri Venkateswara swami. While the experience was spiritually rewarding, the time and effort required to make the trip was making it difficult for many to have his darshan as often as their hearts desired. Many felt a void in their spiritual lives and their discontent was wide spread.
The state of affairs continued until 1985. As a result of philanthropy of nine families who donated 20 acres of land with a farm house in Aurora (a far west suburb of Chicago), as a site for the proposed Balaji temple, a beginning was made in the realization of a dream of many. Since Sri Venkateswara swami (Balaji) temple is an institution that has to serve innumerable generations to come, plans were drawn on a magnificent scale related to the means available and the needs of the moment. No pains were spared in making its designs meet the agamic space requirements when at the same time ensuring that it also meets the local building codes, structural safety standards and the needs of the community.
Padmasri M.Muthiah Sthapathy, a noted expert on temple construction in India collaborated with Sri Subhash Nadkarni, a reputed architect of Chicago and a devotee of Balaji, and together they designed the temple. This design is an excellent blend of ancient Shilpa shastra and modern architectural technology.
Sri Venkateswara Swami Temple of Greater Chicago Aurora This design also served as a model for other temples that were built later in the United States. A unique feature of the temple is provision for circumambulation (pradakshina) inside the temple not only around the presiding deity, but around all the major shrines inside the temple.
Members in several other hindu temples across North America and India graciously shared their experience and knowledge on various temple construction related aspects. Their guidance has been very insightful in helping shape our execution plan. Selection of location, cost involved in construction at various locations that were considered were analyzed using facts from engineers.
Local engineers and architects expedited initial drafts with their relevant expertise. Such strong support from local talent will be essential to our timely success. Active community involvement helped in the diligent efforts towards seeking multiple alternatives for key construction related decisions. Process for Sthapathi selection, master plan input, fund raising schemes, and alignment with Agamic traditions are a few such decisions. Enthusiastic community support helped bolster the fund raising efforts.
To reinforce the significance and power of community involvement in achieving the common goal, here is a quote from The Dalai Lama – "Without the human community, one single human being cannot survive".


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Share this Story: Driver eyes suing Uber after crash
Driver eyes suing Uber after crash
Maryam Shah
Aug 18, 2015 • August 18, 2015 • 2 minute read
Photo by Jack Boland/Toronto Sun
Tawfiqul Alam says he was only a few weeks into his new gig as an UberX driver when a crash totalled his minivan and left him injured and uninsured.
The 48-year-old father of two says he's in pain, unable to work, and considering suing Uber.
"The situation is so frustrating," Alam said Tuesday.
Alam's insurance company, The Co-operators, "denied all benefits based on the fact that he was using the car for commercial use, not personal use," Alam's lawyer Isaac Zisckind said.
He said a vehicle struck Alam's minivan June 9 while he was driving a customer.
Alam sustained soft tissue injuries to his upper body, neck, and back, and suffered psychological distress, Zisckind said.
He said Alam requires physiotherapy and has taken loans to stay afloat.
"He didn't know how complicated it was going to be and how his insurance company would basically leave him stranded," the lawyer said. "What upsets him the most is that Uber really hasn't taken a position. They haven't said, 'Don't worry, we'll take care of it.'"
UberX drivers are not required to buy commercial insurance and Uber has repeatedly said in the past its drivers are covered by $5-million liability insurance for injury and property damage.
Uber spokesman Susie Heath wouldn't discuss Alam's case in detail Tuesday.
"While I can't share specifics on a particular incident, I can tell you that all UberX rides are insured with liability coverage," she wrote. "Our insurer reviews these claims and we are committed to helping drivers understand the resolution process."
Uber's website says "in the event of an accident during an UberX trip, passengers, pedestrians, other drivers, and the community at large can rest assured knowing that ridesharing partners are well covered by commercial auto insurance in addition to any insurance coverage maintained by the driver."
The Co-operators also wouldn't address Alam's specific case.
Spokesman Leonard Sharman said standard personal auto insurance in Ontario doesn't cover a vehicle used to transport paying passengers.
"Any driver who is carrying paying passengers should disclose that to their auto insurer for this reason," he wrote.
Before he began driving, Alam says Uber assured him, "You're good to go."
"There's no indication that a commercial policy is needed," his lawyer said.
maryam.shah@sunmedia.ca
Driver eyes suing Uber after crash Back to video


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Tyson Announces New 1,600-Job Plant
The company plans to invest $320 million in the plant, and complete it by mid-2019.
TONGANOXIE, Kan. (AP) — Tyson Foods Inc. will invest $320 million in a new chicken-processing plant in northeastern Kansas employing 1,600 people so that it can keep up with a growing consumer demand for fresh poultry, the company and state officials announced Tuesday.
The Springdale, Arkansas-based meat producer unveiled its plans during a news conference in Tonganoxie with Gov. Sam Brownback and other state and local officials. The company plans to build the plant outside the town of around 5,300 residents about 30 miles (48 kilometers) west of Kansas City and expects it to be open in mid-2019.
Kansas offered the company a package of incentives "appropriate to a project of this size" to attract the plant, said Kevin Doel, a spokesman for the state Department of Commerce. He declined to provide details, noting that the department does not release such information until contracts with the state are signed.
"We certainly needed to bring something to the table," Doel said. "We were competing with other states."
But the announcement also attracted a small protest from a few local residents who suggested the company would provide low-wage jobs, the Lawrence Journal-World reported. Tyson said jobs would have starting wages of $13 to $15 an hour.
Stephanie Ventura and her husband relocated to Tonganoxie about three years ago from Emporia, where Tyson operates a beef-processing plant. She does not look forward to having Tyson move in.
"It stinks up the whole town," she said. "It's just not a wonderful way of life."
The company plans to buy a 300-acre site and also build a hatchery and feed mill. It expects to break ground on the project this fall.
"We believe this new operation, which will incorporate the latest production technology, will enable us to meet the sustained growth in consumer demand for fresh chicken," Tyson President and CEO Tom Hayes said in a statement.
The plant will be able to process 1.25 million birds a week, and the company plans to contract with local farmers. It and state and local officials project that the plant's payroll, payments to farmers and purchases of feed and utilities will total $150 million a year.
"The far-reaching impact of this development will be felt by farmers, ranchers, agribusinesses and communities throughout eastern Kansas," Brownback said in a statement.
Tyson already operates six plants in Kansas that employ about 5,700 workers and made more than $2 billion last year in payments to cattle and hog suppliers. The company employs about 114,000 workers total and is one of the world's largest food companies.
Doug Ramsey, the company's group president for poultry, said eastern Kansas is "the right location" for a new plant because of the availability of grain and labor and because the state has a "top-notch" transportation system.
Tyson: Price-Fixing Probe is Closed
Tyson Announces SEC Investigation
Tyson CEO to Step Down
Nov 21st, 2016


Urge your Representative to support the No Taxpayer Funding for Abortion Act.
From National Right to Life. The No Taxpayer Funding for Abortion Act (H.R. 7)would basically establish a permanent, government-wide ban on federal subsidies for abortion.
Click here to read about the bill, and take action by entering your zip code in the "Take Action Now" box. You will then have to enter your contact information and write your own letter. A simple "Please support H.R. 7, the No Taxpayer Funding for Abortion Act" will do, but feel free to add more of your own words.
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Die Tous-Talsperre liegt 45 km südwestlich von Valencia in der Nähe der Gemeinde Tous und staut den Júcar auf. Am 20. Oktober 1982 wurde die Talsperre nach extremen Niederschlägen überströmt, worauf der Staudamm brach und eine katastrophale Flutwelle verursachte, bei der es zwischen 20 und 40 Tote gab.

Bau der Talsperre
Mit dem Bau der Talsperre wurde 1957 begonnen. Geplant war zunächst eine 110 m hohe Gewichtsstaumauer. Zwischen 1964 und 1974 gab es wegen geologischer Störungen im Baugrund einen Baustopp, weil sich die Mauer bei dem vorhandenen Baugrund als nicht tauglich erwies. Nach Planungsänderungen wurde die Talsperre 1980 mit einem Staudamm im Mittelstück fertiggestellt. Das Bauwerk war jetzt eine Kombination aus einer Gewichtsstaumauer mit einem Damm mit Lehmkern. Die Talsperre sollte in einer zweiten Bauphase später noch erhöht werden.

Die Katastrophe
Einen Tag vor der Katastrophe begannen außergewöhnlich starke Niederschläge im Einzugsgebiet. Das Hochwasser wurde so stark, dass die Talsperre sehr schnell eingestaut wurde. Zeitweise flossen der Talsperre ca. 5000 m³/s zu. Der elektrische Strom, Telefon und Notstromaggregate fielen aus. Alle Grundablässe ließen sich nicht öffnen, so dass die Talsperre nicht entlastet werden konnte. An der Hochwasserentlastung gab es drei Gleitschütze, die normalerweise zum Öffnen hochgezogen werden konnten, aber auch sie funktionierten ohne Stromversorgung nicht und konnten auch manuell nicht betätigt werden. Ein Ersatz-Notstromaggregat konnte nicht schnell genug beschafft werden. Der Wasserstand stieg deswegen stark an. Ab 17:30 Uhr wurde die Dammkrone überflutet.
Schließlich wurde um 18:15 Uhr der Staudammteil einen Meter hoch überströmt, so dass er weggespült wurde. Um 19:15 Uhr brach die Wand zwischen dem Damm und der Hochwasserentlastung zusammen. Am folgenden Tag gegen 10 Uhr war der Stausee ausgelaufen.
Durch die entstehende Flutwelle wurden die Städte Sumacárcer, Alzira und weitere Orte überflutet. Nach verschiedenen Angaben ertranken 20 bis 40 Menschen.

Ursachen

Geologische Schwierigkeiten waren bereits in der Planung bekannt, waren aber keine Ursache für den Bruch. Die Entlastungsorgane einschließlich der Gleitschütze der Hochwasserentlastung haben aus technischen und organisatorischen Gründen versagt, so dass die Talsperre überströmt werden konnte. Es gab somit keinen ausreichend bemessenen Überlauf, der solch große Wassermengen hätte abführen können.

Die neue Talsperre
Von 1990 bis 1996 wurde eine neue größere Talsperre errichtet, die von zentraler Bedeutung für den Hochwasserschutz am Júcar ist. Der Steinschütt-Staudamm hat einen Lehmkern. Er wurde an dem Ort der alten Talsperre errichtet.

Literatur
 W. Nelihsen, I. Silva: Versagen der Talsperre Tous, Wasserbau-Seminar Aachen 1984, ,
 F. Alrudo, J. Mulet: Description of the Tous Dam break case study (Spain).  Journal of Hydraulic Research, Band 45 Extra-Ausgabe, 2007, S. 45–57

Siehe auch
Liste der Talsperrenkatastrophen

Weblinks
 Dammbruch von Tous
 Ryan Schoolmeesters: Case Study: Tous Dam (Spain, 1982) auf damfailures.org

Stausee in Europa
Talsperre in Spanien
Stausee in Spanien
Stauanlagenunfall
Erbaut in den 1980er Jahren
Erbaut in den 1990er Jahren
Katastrophe 1982
Bauwerk in der Provinz Valencia
Geographie (Valencianische Gemeinschaft)
SToustalsperre

redpajama_set_name: RedPajamaWikipedia


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redpajama_set_name: RedPajamaGithub


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School of Music & Media Announces Final Business Case
By Alec Tudor	 May 22, 2019 No Comments
An email was sent out yesterday to the students of the School of Music & Media, regarding the business changes that are to take place over the summer.
The changes involve staff cuts, restructuring and discontinuation of certain modules and programmes, potential developments such as a Pontio Arts collaboration, summer schools and renewed strategies towards research, student engagement and recruitment.
The school's revised business plan also describes a focus on maintaining its strong reputation in research and impact.
The final plan comes as a result of a months-longs consultation period, during which course representatives and Undeb Bangor have offered student input and feedback, which have been taken into account by the school.
Significant changes within the school
A balance between theory and creative activity is to be retained, as creative assessment can potentially be offered within critical modules.
Staff cuts are to be initiated in Music areas such as Ethnomusicology, Performance and Composition (2.8 FTE – Lecturer level), and in the area of Media Studies (0.5 FTE – Senior Lecturer level).
The variety of Music modules will be considerably reduced, with still enough choice for students.
Students will potentially be able to benefit from new opportunities provided by a collaboration with Pontio Arts.
The creative writing branch of the school will be moved to the School of Languages, Literatures and Linguistics. Journalism and Screenwriting, however, will remain in the school.
The MA Filmmaking course is also to be discontinued in 2020, resulting in a further cut of 0.6 FTE – Senior Lecturer level.
A range of new programmes for 2019-2020 and 2020-2021 is considered in order to attract more students.
A two-year accelerated Music single honours programme possibly to be available at the start of the 2020-2021 academic year.
An MA in Creative Studies is to be explored.
A long term potential to develop Drama as a programme is also considered, but with further development in Music Technology refused.
There has been a negative response concerning the integration of production modules into critical theory ones.
"The school is placing academia over practice when practice informs the theory," said Hannah Grimston, Third Year Creative Studies student.
"I think that this information has been sent out at an extremely damaging time when many students are either away or finished meaning they will not read their emails, furthermore the actual information is hidden without a business proposal rather than being labelled clearly.
I think this decision is offensive to not only the hardworking staff of the school but also to the current and future students attending the university."
Further statements
Bangor University is still to remain the only in Wales to offer a music degree programme completely taught through the Welsh language.
The increase in the number of international students, along with further partnerships and growth plans are considered to be enough to prevent any further cuts to the school.
The school also hopes to maintain its strong research and teaching reputation, stating that the job cuts and redundancies are to be considered carefully.
The School of Music & Media is one of the schools that have been affected by the £5 million cut plan launched by the University in October. It comes after the institution's decision to close the School of Chemistry, and after a previous change in leadership.
Alec Tudor
Editor-in-Chief | 19-20 Creative Corner Editor | 18-19
President Mark Barrow – Two years with Undeb Bangor June 26, 2020
Young artists fight back against government… October 10, 2020


Authors: NPS-CHDN, Richard Worthington, et al.
Abstract: This is a species list for Guadalupe Mountains NP being developed by the Chihuahuan Desert Network (CHDN). This list is situated within the entire Guadalupe Mountains flora that is an ongoing project of Richard Worthington and the Floristic Inventories of the Southwest Project at UTEP. There is substantial overlap between the two floras, but this list represents a more expansive flora than has been documented within the boundaries of the park.


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Clarke leads Australia to their best day of the Ashes
It took them five innings to do so, but Australia finally had a dominant outing with the bat on the first day of the third Ashes test at Old Trafford. My reflections on a day of cricket played out on a batting paradise include:
Shane Watson's sedate 19 proved that you can't give away an inch in test cricket. It was clear from the offset that Watson was going to put in a concerted effort to avoid getting out leg before on the front foot. In order to rectify the issue that has dogged him all series, Watson attempted to play every ball late, even reigning in his natural attacking game. Watson could be forgiven for thinking he had seen out the threat of the new ball before Tim Bresnan delivered the perfect outswinger to snare him in the slips.
While Chris Rogers hasn't been a runaway success this series, he has at least displayed the requisite technique and resolve to survive in the top order unlike many of his compatriots. At Old Trafford, Rogers shifted gears and played a very fluent knock. In the first session of the match Rogers drove excellently, with some beautiful straight drives past the fast bowlers particularly delightful to watch. Ideally, Rogers should have scored a century, but his 84 did end up giving Australia a strong platform to build a big first innings score.
The DRS has grabbed far too many headlines in this series, and Old Trafford also witnessed a terrible howler. Once again, the fault isn't with the system, but with human error. Usman Khawaja was given out caught behind to Graeme Swann by Tony Hill. Khawaja immediately asked for a review, and Kumara Dharmasena in the third umpire's box unbelievably decided against overturning the decision. The slow-motion replay clearly illustrated a gap between bat and ball, Hot Spot didn't reveal an edge and there was barely any sound. There has been some really poor umpiring on display by the third umpires in this series, and for the sake of the cricket fan, I am hoping against hope for an upturn in the standard of decision making from here on in.
A 'Captain's Innings' is probably the most ubiquitous refrain from commentators and writers alike whenever a captain scores a century. In the context of this series, Michael Clarke's unbeaten 124 truly was a captain's innings. Clarke came in after the unjust dismissal of Khawaja, and it would have been very easy for the Australian batsmen to feel sorry for themselves and go into a collective meltdown. In stead, Clarke played some beautiful shots to start his innings at a good pace, and then hatched down for the long haul. He was unruffled throughout his stay at the crease, didn't offer a single chance, and inevitably compiled a very good century. In addition to the scoreboard, Clarke's innings will surely have given the Australian team a shot of confidence in what has been a difficult tour.
The benefits of a good run-out in a warm-up game could be seen in Steven Smith's unbeaten 70. Yes, Smith conceivably could have been dismissed four times during his stay at the crease. Nonetheless, he made the best of the fortune bestowed upon him, diligently leaving balls outside the off stump, and capitalizing on anything short and on his legs. I remarked after the first innings of the first test, that Smith has displayed a level-headed temperament in recent series, and he has now surely sealed his place in the Australian middle order for the rest of this series as well as the return series in Australia later this year.
The 174 runs that Clarke and Smith have compiled together is easily the best batting partnership for Australia this series. Both played with maturity, but the real hallmark of the partnership was the excellent running between the wickets. They turned many ones into twos, and many twos into threes, running England's fielders ragged and crucially denying England's bowlers any chance at establishing a rhythm by constantly rotating the strike.
Stuart Broad was extremely unlucky not to have Smith's wicket just after tea. Broad had Smith plumb in front of the wicket, but with no reviews remaining for England, Smith escaped and potentially changed the course of the match. A return of 0 for 80 reflects poorly on Broad's efforts, as he often bowled in the corridor of danger. One gets the feeling that Broad isn't far away from taking a significant haul of wickets, all he needs is a couple of edges to go his way, and England's ultimate confidence man could become a major player in the series.
England certainly can't control the weather, but they can control the condition of the pitches in a home series. While Australia are certainly at their weakest on slow low pitches, there are days when the English bowling attack can look extremely pedestrian on such wickets. Of course England are pinning their hopes on Swann exploiting conditions on the last two days of the match, but even the world's premier off spinner needs the pitch to have pace in order to be at his most dangerous. Considering Australia's weakness against swing is just as glaring as it is against spin, England could do worse than cater to their three fast bowlers.
That being said it was no secret beforehand that Old Trafford was going to be slow and low. Keeping that in mind, it would have been nice to see England abandon their textbook formulas for once by selecting Monty Panesar ahead of Tim Bresnan to provide some variety to their attack.
So how do Australia consolidate their position? Crossing 300 for the first time in these Ashes must feel like an accomplishment, but they certainly can't afford to rest on their laurels. Clarke and Smith should first and foremost see out the first hour and possibly the whole session. If Australia can get to 400 with the loss of only four wickets, they can really push on aggressively against an English attack that really doesn't relish bowling when the opposition is primed for an assault.
We can still win the Ashes, says Michael Clarke
Labels: Cricket, Graeme Swann, Michael Clarke, Old Trafford, Stuart Broad, Test cricket, Tim Bresnan, Usman Khawaja


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CERTIFICATIONS OF CHIEF EXECUTIVE OFFICER AND CHIEF FINANCIAL OFFICER
PURSUANT TO SECTION 906 OF THE SARBANES-OXLEY ACT OF 2002
(18 U.S.C. SECTION 1350)
I, Stephen E. Croskrey, certify, as of the date hereof, pursuant to 18 U.S.C. Section 1350, as adopted pursuant to Section 906 of the Sarbanes-Oxley Act of 2002, that this Quarterly Report of Danimer Scientific, Inc. on Form 10-Q for the fiscal quarter ended March 31, 2021 fully complies with the requirements of Section 13(a) or 15(d) of the Securities Exchange Act of 1934 and that information contained in such Form 10-Q fairly presents in all material respects the financial condition and results of operations of Danimer Scientific, Inc. at the dates and for the periods indicated.
I, John A. Dowdy, III, certify, as of the date hereof, pursuant to 18 U.S.C. Section 1350, as adopted pursuant to Section 906 of the Sarbanes-Oxley Act of 2002, that this Quarterly Report of Danimer Scientific, Inc. on Form 10-Q for the fiscal quarter ended March 31, 2021 fully complies with the requirements of Section 13(a) or 15(d) of the Securities Exchange Act of 1934 and that information contained in such Form 10-Q fairly presents in all material respects the financial condition and results of operations of Danimer Scientific, Inc. at the dates and for the periods indicated.
/s/ John A. Dowdy, III
John A. Dowdy, III
(Principal Financial Officer)
A signed original of this written statement required by Section 906 has been provided to Danimer Scientific, Inc. and will be retained by Danimer Scientific, Inc. and furnished to the Securities and Exchange Commission or its staff upon request.