text:
\begin{document}

\title{Spatial pseudoanalytic functions arising from the factorization of linear
second order elliptic operators }
\author{Vladislav V. Kravchenko$^{1}$, S\'{e}bastien Tremblay$^{2}$\\$^{1}$Departamento de Matem\'{a}tica, CINVESTAV del IPN, Unidad\\Quer\'{e}taro, Libramiento Norponiente No.~2000 C.P. 76230 Fracc.\\Real de Juriquilla, Quer\'{e}taro, Mexico\\$^{2}$D\'{e}partement de math\'{e}matiques et d'informatique, Universit\'{e} du\\Qu\'{e}bec, Trois-Rivi\`{e}res, Qu\'{e}bec, G9A 5H7, Canada}
\maketitle

\begin{abstract}
Biquaternionic Vekua-type equations arising from the factorization of linear
second order elliptic operators are studied. Some concepts from classical
pseudoanalytic function theory are generalized onto the considered spatial
case. The derivative and antiderivative of a spatial pseudoanalytic function
are introduced and their applications to the second order elliptic equations
are considered.

\end{abstract}

\section{Introduction}

Theory of pseudoanalytic functions \cite{Berskniga}, \cite{Vekua} is one of
the classical branches of complex analysis extending the concepts and ideas from
analytic function theory onto a much more general situation and involving
linear elliptic equations and systems with variables coefficients. Its
development in forties-fifties of the last century was fast and deep and
historically represented an important impulse to the progress in the general
theory of elliptic systems. Nevertheless the important obstacles for a further
development of pseudoanalytic function theory were its limited practical
applications together with the fact that many important results remained in
the level of existence without a possibility to make them really applicable
for solving problems of mathematical physics. Recent progress in
pseudoanalytic function theory reported in \cite{APFT} shows its deep relation
to the stationary Schr\"{o}dinger equation and more general linear second
order elliptic equations and includes new results which allow one to make the
basic objects of the theory and their applications fully explicit. Among other
results it is worth mentioning the possibility to obtain complete systems of
solutions to second order elliptic equations with variable coefficients and
their use for solving related boundary and eigenvalue problems \cite{CKR}.

The generalization of the concept of a pseudoanalytic function onto the case
of three or more dimensions was a subject of considerable efforts (see, e.g., \cite{APFT},
\cite{Berglez}, \cite{Mal98}, \cite{KR2010}, \cite{Sprossig2001}
). The development of a spatial generalization of the theory of pseudoanalytic
functions seems to be possible if only the study is restricted to linear first
order systems of relatively special forms otherwise it is difficult to expect
an efficient generalization of the very first concepts of the theory like
derivative and antiderivative not to mention formal powers, the Cauchy
integral formula and other notions. In the present work we study the Vekua
equation arising from the factorization of the stationary Schr\"{o}dinger
operator. This biquaternionic Vekua equation was first introduced in
\cite{K2006}, \cite{APFT} as a direct generalization of the so-called main
Vekua equation closely related to the two-dimensional stationary
Schr\"{o}dinger equation. We show that it is possible in a natural way to
introduce the concept of a derivative for solutions of the biquaternionic main
Vekua equation and exactly as in the two-dimensional case the derivative is a
solution of another Vekua equation the operator of which is involved in the
factorization of the stationary Schr\"{o}dinger operator. Moreover, we also
show that this operation of differentiation is invertible and introduce the
notion of the antiderivative. Among other applications these constructions
allow us to construct exact solutions to the stationary Schr\"{o}dinger
equation from the known ones possessing certain symmetries.

In this work we consider the elliptic case only, however the generalizations
of the presented results onto the hyperbolic situation (see \cite{KRT} and
\cite{KR2010}) are also possible.

\section{Some definitions and results from two-dimensional pseudoanalytic
function theory}

In this section we give some preliminary definitions and results from the
classical (complex) pseudoanalytic function theory. For more details, see
\cite{Berskniga, APFT, BersStat}.

Let $\Omega$ be a simply connected domain in
$\mathbb{R}^{2}$. We say that a pair of complex functions $F$ and $G$
possessing in $\Omega$ partial derivatives with respect to the real variables
$x$ and $y$ is a generating pair if it satisfies the inequality
\[
\operatorname{Im}(\overline{F}G)>0\qquad\text{in }\Omega.
\]
From this definition it follows that if $W:\Omega\rightarrow\mathbb{C}$, then
there exist unique real valued functions $\phi,\psi$ such that
\[
W(z)=\phi(z)F(z)+\psi(z)G(z),\ \ \ \forall z\in\Omega,
\]
where $\phi=\displaystyle \frac{\mathrm{Im}(\overline{W}G)}{\mathrm{Im}
(\overline{F}G)}$ and $\psi=-\displaystyle \frac{\mathrm{Im}(\overline{W}
F)}{\mathrm{Im}(\overline{F}G)}$.

For a given generating pair $(F,G)$ in $\Omega$, we can define an
$(F,G)$-derivative $\dot{W}(z_{0})$ on $W:\Omega\rightarrow\mathbb{C}$ if the
(finite) limit
\[
\dot{W}(z_{0})=\lim_{z\rightarrow z_{0}}\frac{{W(z)-\phi(z_{0})F(z)-\psi
(z_{0})G(z)}}{{z-z_{0}}}
\]
exists. Sometimes instead of the notation $\dot{W}$ for the $(F,G)$-derivative
of $W$ we will use the notation $\displaystyle\frac{d_{(F,G)}W}{dz}$.

We denote by $\partial_{\overline{z}}=\frac{1}{2}(\partial_{x}+i\partial_{y})$
and $\partial_{z}=\frac{1}{2}(\partial_{x}-i\partial_{y})$, where
$\partial_{x}:=\frac{\partial}{\partial x}$ and $\partial_{y}:=\frac{\partial
}{\partial y}$. The following expressions are known as characteristic
coefficients of the pair $(F,G)$
\[
a_{(F,G)}=-\frac{\overline{F}G_{\overline{z}}-F_{\overline{z}}\overline{G}
}{F\overline{G}-\overline{F}G},\qquad b_{(F,G)}=\frac{FG_{\overline{z}
}-F_{\overline{z}}G}{F\overline{G}-\overline{F}G},
\]

\[
A_{(F,G)}=-\frac{\overline{F}G_{z}-F_{z}\overline{G}}{F\overline{G}
-\overline{F}G},\qquad B_{(F,G)}=\frac{FG_{z}-F_{z}G}{F\overline{G}
-\overline{F}G},
\]
where the subindex $\overline{z}$ or $z$ means the application of
$\partial_{\overline{z}}$ or $\partial_{z}$ respectively.

\begin{theorem}
Let $(F,G)$ be a generating pair in $\Omega$ and $W\in C^{1}(\Omega)$. The
$(F,G)$-derivative $\dot{W}=\displaystyle\frac{d_{(F,G)}W}{dz}$ of $W$ exists
and has the form
\begin{equation}
\dot{W}=\phi_{z}F+\psi_{z}G=W_{z}-A_{(F,G)}W-B_{(F,G)}\overline{W}
\label{derivative}
\end{equation}
if and only if
\begin{equation}
W_{\overline{z}}=a_{(F,G)}W+b_{(F,G)}\overline{W}. \label{VekuaGen}
\end{equation}

\end{theorem}

Equation (\ref{VekuaGen}) is called Vekua equation (sometimes Carleman-Vekua
equation) and can also be written in the following form
\[
\phi_{\overline{z}}F+\psi_{\overline{z}}G=0.
\]
The Vekua equation (\ref{VekuaGen}) represents a generalization of the
Cauchy-Riemann system and solutions of this equation are called $(F,G)$
-pseudoanalytic functions. It is easy to see that functions $F$ and $G$ are
$(F,G)$-pseudoanalytic, and $\overset{\cdot}{F}\equiv\overset{\cdot}{G}
\equiv0$.

Let $(F,G)$ and $(F_{1},G_{1})$ be two generating pairs in $\Omega$. We say
that $(F_{1},G_{1})$ is a successor of $(F,G)$ and $(F,G)$ is a predecessor of
$(F_{1},G_{1})$ if
\[
a_{(F_{1},G_{1})}=a_{(F,G)}\qquad\text{and}\qquad b_{(F_{1},G_{1})}
=-B_{(F,G)}\text{.}
\]
The importance of this definition becomes obvious from the following result.

\begin{theorem}
\label{ThBersDer}Let $W$ be an $(F,G)$-pseudoanalytic function and let
$(F_{1},G_{1})$ be a successor of $(F,G)$. Then $\overset{\cdot}{W}$ is an
$(F_{1},G_{1})$-pseudoanalytic function.
\end{theorem}

In other words, when $W$ is an $(F,G)$-pseudoanalytic function then $\dot{W}$
satisfies the Vekua equation $(\dot{W})_{\overline{z}}=a_{(F,G)}\dot
{W}-B_{(F,G)}\overline{\dot{W}}$.

For a given generating pair $(F,G)$ in $\Omega$, the adjoint generating pair
$(F,G)^{\ast}=(F^{\ast},G^{\ast})$ is defined by the formulas
\[
F^{\ast}=-\frac{2\overline{F}}{F\overline{G}-\overline{F}G}\qquad\text{ and }
\qquad G^{\ast}=\frac{2\overline{G}}{F\overline{G}-\overline{F}G}.
\]

The $(F,G)$-integral is defined as follows
\[
\int_{\Gamma}Wd_{(F,G)}z= F(z_{1})\operatorname{Re}\int_{\Gamma}G^{\ast
}Wdz+G(z_{1})\operatorname{Re}\int_{\Gamma}F^{\ast}Wdz
\]
where $\Gamma$ is a rectifiable curve leading from $z_{0}$ to $z_{1}$.

If $W=\phi F+\psi G$ is an $(F,G)$-pseudoanalytic function where $\phi$ and
$\psi$ are real valued functions then
\[
\int_{z_{0}}^{z}\overset{\cdot}{W}d_{(F,G)}z=W(z)-\phi(z_{0})F(z)-\psi
(z_{0})G(z), \label{FGAnt}
\]
and as $\overset{\cdot}{F}=\overset{}{\overset{\cdot}{G}=}0$, this integral is
path-independent and represents the $(F,G)$-antiderivative of $\overset{\cdot
}{W}$.

Consider the equation
\[
\varphi_{z}=\Phi
\]
for a real-valued function $\varphi$ in a simply connected domain, where $\Phi=\Phi_{1}+i\Phi
_{2}$ is a given complex valued function such that its real part $\Phi_{1}$
and imaginary part $\Phi_{2}$ satisfy the equation
\begin{equation}
\partial_{y}\Phi_{1}+\partial_{x}\Phi_{2}=0, \label{casirot}
\end{equation}
then we can reconstruct $\varphi$ up to an arbitrary real constant $c$ in the
following way

\begin{equation}
\varphi(x,y)=2\left(  \int_{\Gamma}\Phi_{1}dx-\Phi_{2}dy\right)  +c.
\label{Antigr2}
\end{equation}

By $A$ we denote the integral operator in (\ref{Antigr2}):
\[
A[\Phi](x,y)=2\left(  \int_{\Gamma}\Phi_{1}dx-\Phi_{2}dy\right)  +c.
\]

Thus if $\Phi$ satisfies (\ref{casirot}), there exists a family of real valued
functions $\varphi$ such that $\varphi_{z}=\Phi$, given by the formula
$\varphi=A[\Phi]$.

In a similar way we define the operator $\overline{A}$ corresponding to
equation $\varphi_{\overline{z}}=\Phi$, where $\Phi=\Phi_{1}+i\Phi_{2}$
satisfies $\partial_{y}\Phi_{1}-\partial_{x}\Phi_{2}=0$, by the following
antiderivative operator:
\[
\overline{A}[\Phi](x,y)=2\left(  \int_{\Gamma}\Phi_{1}dx+\Phi_{2}dy\right)
+c.
\]

Now, it is well known that if $f_{0}$ is a non vanishing particular solution of the
one-dimensional stationary Schr\"{o}dinger equation $\big(-\frac{d^{2}}
{dx^{2}}+q(x)\big)f(x)=0$, the Schr\"{o}dinger operator can be factorized as
$\frac{d^{2}}{dx^{2}}-q=(\frac{d}{dx}+\frac{f_{0}^{\prime}}{f_{0}})(\frac
{d}{dx}-\frac{f_{0}^{\prime}}{f_{0}})$. A similar factorization can be
obtained in the two-dimensional case. First, we define the Vekua operators
\[
V:=\partial_{\overline{z}}-\frac{f_{\overline{z}}}{f}C\quad\text{and}\quad
V_{1}:=\partial_{\overline{z}}+\frac{f_{z}}{f}C,
\]
where $f\in C^{2}(\Omega;\mathbb{R})$ is a nonvanishing real-valued function
defined in $\Omega$ and $C$ is the complex conjugate operator. The
two-dimensional stationary Schr\"{o}dinger equation
\begin{equation}
\big(-\triangle+q(x,y)\big)\varphi=0\qquad\text{ in }\Omega, \label{Schro1}
\end{equation}
where $\triangle$ is the two-dimensional Laplacian $\triangle:=\frac
{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}$, $q$ and
$\varphi$ are real-valued functions, $\varphi\in C^{2}(\Omega;\mathbb{R})$,
can be factorized as follows.

\begin{theorem}
\emph{\cite{K2005}} Let $f$ be a real-valued nonvanishing solution in $\Omega$
of the Schr\"{o}dinger equation (\ref{Schro1}). Then for any real-valued
function $\varphi\in C^{2}(\Omega)$ the following equalities hold
\begin{equation}
\frac{1}{4}(\triangle-q)\varphi= V_{1}\overline{V}\varphi=\overline{V}
_{1}V\varphi. \label{factorization}
\end{equation}

\end{theorem}

\begin{theorem}
\emph{\cite{K2006}} Let $W=W_{1}+iW_{2}$ be a solution of the Vekua equation
$VW=0$. Then $W_{1}=\text{\emph{Re} }W$ is a solution of the Schr\"{o}dinger
equation (\ref{Schro1}) in $\Omega$ and $W_{2}=\text{\emph{Im} }W$ is a
solution of the associated Schr\"{o}dinger equation
\begin{equation}
(-\triangle+r)\psi=0\quad\text{in }\Omega,\quad\quad r=2\left(  \frac{\nabla
f}{f}\right)  ^{2}-q, \label{Schro2}
\end{equation}
where $(\nabla f)^{2}=f_{x}^{2}+f_{y}^{2}$. \label{VekuasolSchr}
\end{theorem}

Theorem \ref{VekuasolSchr} tells us that as much as real and imaginary parts
of a complex analytic function are harmonic functions, the real and imaginary
parts of a solution of the Vekua equation $VW=0$ are solutions of associated
stationary Schr\"{o}dinger equations. Now, we know that given an arbitrary
real-valued harmonic function in a simply connected domain, a conjugate
harmonic function can be constructed explicitly such that the obtained pair
of harmonic functions represent the real and imaginary parts of a complex
analytic function. This corresponds to the more general situation for
solutions of associated stationary Schr\"{o}dinger equations; the following
theorem gives us the precise result.

\begin{theorem}
\emph{\cite{K2005}\label{ThConjHarm}} Let $W_{1}$ be a real-valued solution of
(\ref{Schro1}) in a simply connected domain $\Omega$. Then the real-valued
function $W_{2}$, a solution of (\ref{Schro2}) such that $W=W_{1}+iW_{2}$ is a
solution of $VW=0$, is constructed according to the formula
\[
W_{2}=f^{-1}\overline{A}[if^{2}\partial_{\overline{z}}(f^{-1}W_{1})].
\]

Given a solution $W_{2}$ of (\ref{Schro2}), the corresponding solution $W_{1}$
of (\ref{Schro1}) such that $W=W_{1}+iW_{2}$ is a solution of $VW=0$, is
constructed as
\[
W_{1}=-f\overline{A}[if^{-2}\partial_{\overline{z}}(fW_{2})].
\]

\end{theorem}

\section{Pseudoanalytic derivative for a class of Vekua equations in the
plane\label{SectTwoDim}}

For $f\in C^{2}(\Omega,\mathbb{R})$, a nonvanishing real-valued function
defined in $\Omega$, it is known \cite[Sect. 3.4]{APFT} that the functions
$F=f$ and $G=i/f$ represent a generating pair for the Vekua equation
\begin{equation}
VW=0, \label{VekuaMain}
\end{equation}
and if one fixes this generating pair $(F,G)=(f,i/f)$, the equation
\begin{equation}
V_{1}w=0 \label{Vekua1}
\end{equation}
is satisfied by the $(F,G)$-derivatives of solutions of (\ref{VekuaMain}) and
in particular by the successor $(F_{1},G_{1})$, and hence it is natural to
call $V_{1}$ a successor of $V$.

Note that from (\ref{derivative}) the $(F,G)$-derivative in this case is given
as
\[
\overset{\cdot}{W}=\frac{d_{(F,G)}W}{dz}=W_{z}-\frac{f_{z}}{f}\overline
{W}=CVCW.
\]
Thus, the fact that $\overset{\cdot}{W}$ is a solution of (\ref{Vekua1}) can
be written in the following form. If $W$ is a solution of (\ref{VekuaMain})
then
\[
V_{1}CVCW=0.
\]
Note the following commutation relation
\[
CV=\overline{V}C
\]
where $\overline{V}=\partial_{z}-\frac{f_{z}}{f}C$. That is,
\[
\overset{\cdot}{W}=\overline{V}W
\]
and $V_{1}\overset{\cdot}{W}=V_{1}\overline{V}W$. The equality to zero of this
last expression in the case when $W$ is a solution of (\ref{VekuaMain}) is a
direct consequence of the factorization (\ref{factorization}), i.e.
$V_{1}\overline{V}\varphi=\frac{1}{4}(\Delta-q)\varphi$ where $q=\frac{\Delta
f}{f}$ and $\varphi$ is real-valued, twice continuously differentiable
function, and
\[
V_{1}\overline{V}(i\phi)=\frac{i}{4}(\Delta-r)\phi
\]
where $r=-q+2\left(  \frac{\nabla f}{f}\right)  ^{2},$ as well as of the fact
that from theorem \ref{VekuasolSchr} if $W$ is a solution of (\ref{VekuaMain})
then $(-\Delta+q)W_{1}=0$ and $(-\Delta+r)W_{2}=0$ where $W_{1}
=\operatorname{Re}W$ and $W_{2}=\operatorname{Im}W$.

Let us notice that the $(F,G)$-derivative of $W$ can be also represented as
follows
\begin{equation}
\overline{V}W=f\partial_{z}(f^{-1}W_{1})+if^{-1}\partial_{z}(fW_{2}).
\label{SecondRep1}
\end{equation}
This form suggests the definition of the inverse operation, the
antiderivative:
\[
w=fA(f^{-1}\Phi_{1})+if^{-1}A(f\Phi_{2})
\]
where $\Phi=\Phi_{1}+i\Phi_{2}$ is any solution of (\ref{Vekua1}). The
resulting function $w$ is necessarily a solution of (\ref{VekuaMain}).

From (\ref{SecondRep1}) it can be seen that the $(F,G)$-derivative can be
written as
\begin{equation}
\overset{\cdot}{W}=\varphi_{z}F+\psi_{z}G \label{SecondRep}
\end{equation}
where $\varphi=f^{-1}W_{1}$ and $\psi=fW_{2}$.

\section{Three-dimensional pseudoanalytic function\newline theory for a class
of Vekua equations}

\subsection{General definitions and results}

We denote by $\mathbb{H}(\mathbb{C})=\{Q\ |\ Q=\sum_{\alpha=0}^{3}Q_{\alpha
}\mathbf{e}_{\alpha}\}$ the algebra of complex quaternions (biquaternions)
(see, e.g., \cite{AQA, IRSM}), where $Q_{\alpha}\in\mathbb{C}$, $\mathbf{e}
_{0}=1$, $\{\mathbf{e}_{k}\ |\ k=1,2,3\}$ are the standard quaternionic
imaginary units satisfying $\mathbf{e}_{\alpha}\mathbf{e}_{\beta}
+\mathbf{e}_{\beta}\mathbf{e}_{\alpha}=-2\delta_{\alpha,\beta}$, where
$\delta_{\alpha,\beta}$ is the usual Kronecker delta. The imaginary unit in
$\mathbb{C}$ is denoted by $i$ as usual and commutes with all basic units
$\mathbf{e}_{\alpha}$, $\alpha=0,1,2,3$. The vector representation of
$Q\in\mathbb{H}(\mathbb{C})$ can also be used: $Q=\text{Sc}(Q)+\text{Vec}(Q)$,
where $\text{Sc}(Q)=Q_{0}$ and $\text{Vec}(Q)=\mathbf{Q}=\sum_{k=1}^{3}
Q_{k}\mathbf{e}_{k}$. The quaternionic conjugation is defined by $\overline
{Q}=Q_{0}-\mathbf{Q}$ (sometimes the quaternionic conjugation operator $C_{H}$
is used, i.e. $C_{H}Q=\overline{Q}$).

By $M^{P}$ we denote the operator of multiplication by a biquaternion $P$ from
the right-hand side, i.e.
\[
M^{P}Q=Q\cdot P.
\]

Let $Q$ be a complex quaternion-valued differentiable function of
$\mathbf{x}=(x_{1},x_{2},x_{3})$. We define the Dirac operator (sometimes
called the Moisil-Theodorescu operator) as
\begin{equation}
DQ=\sum_{k=1}^{3} \mathbf{e}_{k}\ \partial_{k} Q, \label{Diracoperator}
\end{equation}
where $\partial_{k}:=\displaystyle \frac{\partial}{\partial x_{k}}$.
Expression (\ref{Diracoperator}) can be rewritten in a vector form as
\[
DQ=-\text{div } \mathbf{Q}+\nabla Q_{0}+\text{rot } \mathbf{Q},
\]
in other words $\text{Sc}(DQ)=-\text{div }\mathbf{Q}$ and $\text{Vec}
(DQ)=\nabla Q_{0} +\text{rot } \mathbf{Q}$. Let us notice that $D^{2}
=-\triangle$, where $\triangle:=\sum_{k=1}^{3} \partial_{k}^{2}$ is the
Laplacian in $\mathbb{R}^{3}$.

We notice that the Dirac operator was introduced as acting from the left-hand
side. The corresponding operator acting from the right-hand side we will
denote by $D_{r}Q=\sum_{k=1}^{3}\partial_{k}Q\ \mathbf{e}_{k}$. In the vector
form $D_{r}$ can be represented as $D_{r}Q=-\text{div }\mathbf{Q}+\nabla
Q_{0}-\text{rot }\mathbf{Q}$.

The Dirac operator admits the following generalization of the Leibniz rule
(see, e.g., \cite{GurlebeckSprossig}):

\begin{theorem}
Let $\{P,Q\}\subset C^{1}\big(\Omega;\mathbb{H}(\mathbb{C})\big)$, where
$\Omega$ is some domain in $\mathbb{R}^{3}$. Then
\[
D[P\cdot Q]=D[P]\cdot Q+\overline{P}\cdot D[Q]+2\big(\text{\emph{Sc}
}(PD)\big)[Q],
\]
where
\[
\big(\text{\emph{Sc}}(PD)\big)[Q]=-\sum_{k=1}^{3} P_{k}\partial_{k} Q.
\]

\end{theorem}

In particular, we observe that when $\text{Vec}(P)=0$, i.e. $P=P_{0}$, then
\[
D[P_{0}\cdot Q]=D[P_{0}]\cdot Q+P_{0}\cdot D[Q].
\]

Consider now the equation
\[
\nabla\varphi=\mathbf{\Psi}
\]
where $\mathbf{\Psi}=\sum_{k=1}^{3} \Psi_{k} \mathbf{e}_{k}$ is a given
complex-valued vector such that $\text{rot }\mathbf{\Psi}\equiv0$. The
complex-valued scalar function $\varphi$ is then said to be the potential of
$\mathbf{\Psi}$. We will write $\varphi=\mathcal{A}[\mathbf{\Psi}]$. The
operator $\mathcal{A}$ is a simple generalization of the opeators $A$ and
$\overline{A}$ of the preceding section. It is well known that we can
reconstruct $\varphi$, up to an arbitrary complex constant $c$, in the
following way
\[
\mathcal{A}[\mathbf{\Psi}](x,y,z)=\int_{x_{0}}^{x}\Psi_{1}(\xi,y_{0}
,z_{0})d\xi+ \int_{y_{0}}^{y}\Psi_{2}(x,\eta,z_{0})d\eta+\int_{z_{0}}^{z}
\Psi_{3}(x,y,\zeta)d\zeta+c,
\]
where $(x_{0},y_{0},z_{0})$ is an arbitrary point in the domain of interest.

Let us now consider the three-dimensional stationary Schr\"{o}dinger equation
\begin{equation}
(-\Delta+q)\phi=0\qquad\text{ in }\Omega, \label{Schro3d}
\end{equation}
where $q$ and $\phi$ are complex-valued functions depending on $\mathbf{x}
=(x_{1},x_{2},x_{3})$ and $\Omega$ is a domain in $\mathbb{R}^{3}$. Using the
Dirac operator, one can factorize this equation in a way similar to that in
theorem~\ref{factorization}.

\begin{theorem}
\emph{\cite[Sect. 16.1]{APFT}} Let $f$ be a nonvanishing particular solution
of (\ref{Schro3d}). Then for any scalar (complex-valued) function $\phi\in
C^{2}(\Omega;\mathbb{C})$ the following equality holds:
\begin{equation}
(-\Delta+q)\phi=(D+M^{\frac{Df}{f}})(D-\frac{Df}{f}C_{H})\phi. \label{Fact}
\end{equation}

\end{theorem}

The equivalent of theorem \ref{VekuasolSchr} in the three-dimensional case is
given here.

\begin{theorem}
\label{ThSolutionsOfVekuaMain3D}Let $W_{0}+\mathbf{W}$ be a solution of the
equation
\begin{equation}
\left(  D-\frac{Df}{f}C_{H}\right)  W=0. \label{mainVekuaquat}
\end{equation}
Then $W_{0}$ is a solution of (\ref{Schro3d}) with $q=\frac{\triangle f}{f}$;
the scalar function $\varphi_{0}=W_{0}/f$ is a solution of the equation
\begin{equation}
\text{\emph{div}}(f^{2}\nabla\varphi_{0})=0 \label{eqphi0}
\end{equation}
and the vector function $\mathbf{\Phi}=f\mathbf{W}$ is a solution of the
equation
\begin{equation}
\text{\emph{rot}}(f^{-2}\text{ \emph{rot} }\mathbf{\Phi})=0. \label{eqPhi}
\end{equation}

\end{theorem}

\subsection{Pseudoanalytic derivative in the space for a class of Vekua
equations}

We denote the operators
\[
V:=D-\frac{Df}{f}C_{H}\quad\text{and}\quad\overline{V}_{1}:=D+M^{\frac{Df}{f}
}\text{.}
\]
Here the notations from Section \ref{SectTwoDim} are preserved in order to
make the analogy between the two-dimensional and the higher dimensional case
more transparent.

Application of the operator $C_{H}$ to (\ref{Fact}) gives us
\[
(-\Delta+q)\phi=(D_{r}+\frac{Df}{f})(D_{r}-M^{\frac{Df}{f}}C_{H})\phi.
\]
We denote
\[
\overline{V}:=D_{r}-M^{\frac{Df}{f}}C_{H}\quad\text{and}\quad V_{1}
:=D_{r}+\frac{Df}{f}.
\]
All the operators $V$, $\overline{V}$, $V_{1}$ and $\overline{V}_{1}$ are
considered in application to $\mathbb{H}(\mathbb{C})$-valued, continuously
differentiable functions. Note that the following relations hold
\[
C_{H}V=-\overline{V}C_{H}\quad\text{and}\quad C_{H}V_{1}=-\overline{V}
_{1}C_{H}.
\]

Suppose
\begin{equation}
VW=0 \label{VekuaMain3D}
\end{equation}
where $W\in C^{1}(\Omega;\mathbb{H}(\mathbb{C}))$. Then due to (\ref{Fact}) we
have
\begin{equation}
\overline{V}_{1}VW_{0}=V_{1}\overline{V}W_{0}=(-\Delta+q)W_{0}=0
\label{factoriz}
\end{equation}
and hence
\[
0=\overline{V}_{1}V\mathbf{W}=C_{H}\overline{V}_{1}V\mathbf{W}=V_{1}
\overline{V}\mathbf{W}.
\]
Thus, from this and from (\ref{factoriz}) the $\mathbb{H}(\mathbb{C})$-valued
function $\overline{V}W$ is a solution of the equation
\begin{equation}
V_{1}w=0 \label{Vekua1_3D}
\end{equation}
whenever $W$ is a solution of (\ref{VekuaMain3D}). Consequently we define the
Bers derivative of a solution $W$ of (\ref{VekuaMain3D}) as follows

\begin{definition}
Let $W$ be a solution of (\ref{VekuaMain3D}). Then the function
\[
\overset{\cdot}{W}=\overline{V}W
\]
is a solution of (\ref{Vekua1_3D}) and is called the derivative of $W$.
\end{definition}

In other words, $\overline{V}:\ker V\rightarrow\ker V_{1}$. Moreover, as we
show below the derivative $\overset{\cdot}{W}$ is purely vectorial, and hence
it is a solution of the equation $\overline{V}_{1}w=0$ as well.

Let us obtain another representation for $\overset{\cdot}{W},$ similar to
(\ref{SecondRep}). We have
\[
\overline{V}W=\left(  D_{r}-M^{\frac{Df}{f}}C_{H}\right)  W=\left(
D_{r}-M^{\frac{Df}{f}}C_{H}\right)  \sum_{\alpha=0}^{3}\varphi_{\alpha
}F_{\alpha}
\]
where $F_{\alpha}$ represent the generating quartet \cite[Sect. 16.2]{APFT}
for equation (\ref{VekuaMain3D})
\[
F_{0}=f,\quad F_{1}=\frac{\mathbf{e}_{1}}{f},\quad F_{2}=\frac{\mathbf{e}_{2}
}{f},\quad F_{3}=\frac{\mathbf{e}_{3}}{f},
\]
$\varphi_{k}$ are scalar functions defined by $\varphi_{0}=W_{0}/f$ and
$\varphi_{k}=fW_{k}$ for $k=1,2,3$. Then
\begin{align*}
\overline{V}W  &  =\sum_{\alpha=0}^{3}F_{\alpha}\cdot D\varphi_{\alpha}
+\sum_{\alpha=0}^{3}\varphi_{\alpha}\cdot D_{r}F_{\alpha}-\sum_{\alpha=0}
^{3}\varphi_{\alpha}\overline{F}_{\alpha}\frac{Df}{f}\\
&  =\sum_{\alpha=0}^{3}F_{\alpha}\cdot D\varphi_{\alpha}+\sum_{\alpha=0}
^{3}\varphi_{\alpha}\cdot\overline{V}F_{\alpha}.
\end{align*}
It is easy to see that $\overline{V}F_{\alpha}=0$ for $\alpha=0,1,2,3$.
Indeed, we have
\[
0=VF_{\alpha}=C_{H}VF_{\alpha}=\overline{V}\overline{F}_{\alpha}
\]
and if $\alpha=0$, then $\overline{V}\overline{F}_{0}=\overline{V}F_{0}=0$,
otherwise $\overline{V}\overline{F}_{k}=-\overline{V}F_{k}=0$ for $k=1,2,3$.
Thus, another representation for the derivative $\overset{\cdot}{W}$ has the
form
\begin{equation}
\overset{\cdot}{W}=\overline{V}W=\sum_{\alpha=0}^{3}F_{\alpha}\cdot
D\varphi_{\alpha}. \label{Derivative}
\end{equation}
Let us obtain the inverse operation,- the antiderivative. We have the
following two equalities
\begin{equation}
\sum_{\alpha=0}^{3}\left(  D\varphi_{\alpha}\right)  F_{\alpha}=0
\label{VekuaSecond}
\end{equation}
and
\begin{equation}
\sum_{\alpha=0}^{3}\left(  D\varphi_{\alpha}\right)  \overline{F}_{\alpha
}=-\overline{\overset{\cdot}{W}}, \label{VekuaDer}
\end{equation}
the first of them being another form of the Vekua equation (\ref{VekuaMain3D})
(see \cite[Sect. 16.2]{APFT}), and the second is obtained from
(\ref{Derivative}) by applying $C_{H}$. More explicitly,
\[
D\varphi_{0}\cdot f+\sum_{k=1}^{3}D\varphi_{k}\cdot\frac{\mathbf{e}_{k}}{f}=0
\]
and
\[
D\varphi_{0}\cdot f-\sum_{k=1}^{3}D\varphi_{k}\cdot\frac{\mathbf{e}_{k}}
{f}=-\overline{\overset{\cdot}{W}}.
\]
Then
\begin{equation}
D\varphi_{0}=-\frac{1}{2f}\overline{\overset{\cdot}{W}}. \label{Dphi0}
\end{equation}
A scalar solution $\varphi_{0}$ of this equation exists if only $\overset
{\cdot}{W}$ is purely vectorial and satisfies the equation
\[
\operatorname*{rot}\left(  \frac{1}{f}\overset{\cdot}{W}\right)  =0.
\]
Consider
\begin{align*}
\overset{\cdot}{W}+\overline{\overset{\cdot}{W}}  &  =\sum_{\alpha=0}
^{3}F_{\alpha}\cdot D\varphi_{\alpha}-\sum_{\alpha=0}^{3}\left(
D\varphi_{\alpha}\right)  \overline{F}_{\alpha}\\
&  =\sum_{k=1}^{3}\frac{\mathbf{e}_{k}}{f}D\varphi_{k}+\sum_{k=1}^{3}
D\varphi_{k}\cdot\frac{\mathbf{e}_{k}}{f}\\
&  =\frac{1}{f}\sum_{k=1}^{3}\left(  \mathbf{e}_{k}D\varphi_{k}+D\varphi
_{k}\mathbf{e}_{k}\right) \\
&  =-\frac{1}{f}\sum_{k=1}^{3}\partial_{k}\varphi_{k}=-\frac{1}{f}
\operatorname*{div}\mathbf{\Phi}
\end{align*}
where $\mathbf{\Phi}:=\sum_{k=1}^{3}\varphi_{k}\mathbf{e}_{k}$.

As $\mathbf{\Phi}=f\mathbf{W}$ we have that \cite[equation (16.11)]{APFT}
\begin{equation}
\operatorname*{div}\mathbf{\Phi}=0. \label{divPhi}
\end{equation}
Thus, $\overset{\cdot}{W}+\overline{\overset{\cdot}{W}}=0$ and hence
$\overset{\cdot}{W}$ is purely vectorial.

Consider the expression
\begin{align*}
\operatorname*{rot}\left(  \frac{1}{f}\overset{\cdot}{W}\right)   &
=\operatorname*{rot}\left(  \frac{1}{f}\sum_{\alpha=0}^{3}F_{\alpha}\cdot
D\varphi_{\alpha}\right) \\
&  =\operatorname*{rot}\left(  \nabla\varphi_{0}+\frac{1}{f^{2}}\sum_{k=1}
^{3}\mathbf{e}_{k}D\varphi_{k}\right) \\
&  =\operatorname*{rot}\left(  \frac{1}{f^{2}}\sum_{k=1}^{3}\mathbf{e}
_{k}D\varphi_{k}\right)  .
\end{align*}
From the previous chain of equalities we already know that
\[
\sum_{k=1}^{3}\mathbf{e}_{k}D\varphi_{k}=-\sum_{k=1}^{3}D\varphi_{k}
\mathbf{e}_{k}=-D\mathbf{\Phi}=-\operatorname*{rot}\mathbf{\Phi},
\]
because of (\ref{divPhi}). Thus,
\[
\operatorname*{rot}\left(  \frac{1}{f}\overset{\cdot}{W}\right)
=-\operatorname*{rot}\left(  \frac{1}{f^{2}}\operatorname*{rot}\mathbf{\Phi
}\right)
\]
and according to theorem \ref{ThSolutionsOfVekuaMain3D} this is zero (see
\cite[theorem 161]{APFT}).

We have then that if $W$ is a solution of (\ref{VekuaMain3D}) then equation
(\ref{Dphi0}) possesses a solution $\varphi_{0}$ which can be written as
follows
\begin{equation}
\varphi_{0}=\frac{1}{2}\mathcal{A}\left[  \frac{1}{f}\overset{\cdot}
{W}\right]  . \label{phi0}
\end{equation}
This expression is unique up to an additive constant.

Now let us see, how $\mathbf{\Phi}$ can be recovered from (\ref{VekuaSecond})
and (\ref{VekuaDer}). We have
\[
\sum_{k=1}^{3}D\varphi_{k}\cdot\frac{\mathbf{e}_{k}}{f}=\frac{1}{2}
\overline{\overset{\cdot}{W}},
\]
that is,
\begin{equation}
D\mathbf{\Phi}=\frac{f}{2}\overline{\overset{\cdot}{W}}. \label{DPhi}
\end{equation}
More explicitly,
\[
\operatorname*{div}\mathbf{\Phi}=0\quad\text{and}\quad\operatorname*{rot}
\mathbf{\Phi=-}\frac{f}{2}\overset{\cdot}{W}.
\]
Notice that $\operatorname*{div} \big(f\overset{\cdot}{W}\big)  =0.$
Indeed, according to (\ref{Dphi0}),
\[
f\overset{\cdot}{W}=2f^{2}\nabla\varphi_{0}=2f^{2}\nabla\left(  \frac{W_{0}
}{f}\right)
\]
and due to theorem \ref{ThSolutionsOfVekuaMain3D}, $\operatorname*{div}\left(
f^{2}\nabla\left(  \frac{W_{0}}{f}\right)  \right)  =0$.

Thus, the problem of recovering $\mathbf{\Phi}$ reduces to the well-studied
problem of reconstruction of a vector by its divergence and rotor.

We will need the following notation. For a given vector function $\mathbf{Q}$,
we define $\mathbf{B}[\mathbf{Q}](\mathbf{x})$ by
\[
\mathbf{B}[\mathbf{Q}](\mathbf{x})=\frac{1}{4\pi}\int_{\Omega}\frac
{\mathbf{Q}(\mathbf{y})}{\left\vert \mathbf{x}-\mathbf{y}\right\vert }
d\Omega,
\]
Note that $\mathbf{B}$ is a right-inverse for the operator
$\operatorname*{rot}\operatorname*{rot}$.

Then we have (see, e.g., \cite[Sect. 5.7]{Korn})
\begin{equation}
\mathbf{\Phi}=\operatorname*{rot}\left(  \mathbf{B}\left[  -\frac{f}{2}\dot
{W}\right]  \right)  +\nabla h, \label{Phi}
\end{equation}
where $h$ is any harmonic function defined in $\Omega$.

These results allow us to formulate the following theorem.

\begin{theorem}
Let $\mathbf{w}\in C^{1}\big(\Omega;\mathbb{H}(\mathbb{C})\big)$ be a purely
vectorial solution of the equation
\begin{equation}
\left(  D+M^{\frac{Df}{f}}\right)  \mathbf{w}=0 \label{D+M}
\end{equation}
where $f$ is a nonvanishing scalar $C^{1}$-function. Then a solution of the
equation
\begin{equation}
DW-\frac{Df}{f}\overline{W}=0\qquad\text{in }\Omega\label{VekuaMain3DExplicit}
\end{equation}
such that $\mathbf{w}=\overset{\cdot}{W}$ is defined as follows
\begin{equation}
W=\frac{1}{2}\left(  f\mathcal{A}\left[  \frac{\mathbf{w}}{f}\right]
-\frac{1}{f}\operatorname*{rot}\left(  \mathbf{B}\left[  f\mathbf{w}\right]
\right)  +\frac{\nabla h}{f}\right)  \label{Antider}
\end{equation}
where $h$ is an arbitrary harmonic function in $\Omega$.
\end{theorem}

\begin{proof}
First, we notice that
\begin{equation}
\operatorname*{rot}\left(  \frac{\mathbf{w}}{f}\right)  =0. \label{rot=0}
\end{equation}
Indeed, from the vector part of (\ref{D+M}) one has $\operatorname*{rot}
\mathbf{w}+(\mathbf{w}\times\frac{\nabla f}{f})=0$ which is equivalent to
(\ref{rot=0}). Thus, the vector $\frac{\mathbf{w}}{f}$ is a gradient and the
expression $\mathcal{A}\left[  \frac{\mathbf{w}}{f}\right]  $ is well defined.

Observe that the operator in equation (\ref{VekuaMain3DExplicit}) can be
written in the following form
\[
D-\frac{Df}{f}C_{H}=\frac{1}{2}\left(  fDf^{-1}(I+C_{H})+f^{-1}Df(I-C_{H}
)\right)
\]
where $I$ is the identity operator. Applying it to (\ref{Antider}) we obtain
\begin{align*}
& \frac{1}{4}\left(  fDf^{-1}(I+C_{H})+f^{-1}Df(I-C_{H})\right)  \left(
f\mathcal{A}\left[  \frac{\mathbf{w}}{f}\right]  -\frac{1}{f}
\operatorname*{rot}\left(  \mathbf{B}\left[  f\mathbf{w}\right]  \right)
+\frac{\nabla h}{f}\right) \\*[2ex]
 = & \frac{1}{2}\left(  fD\mathcal{A}\left[  \frac{\mathbf{w}}{f}\right]
-f^{-1}D\operatorname*{rot}\left(  \mathbf{B}\left[  f\mathbf{w}\right]
\right)  \right) =\frac{1}{2}\left(  \mathbf{w-w}\right)  =0.
\end{align*}
Next we prove that $\mathbf{w}=\overset{\cdot}{W}$. For this we observe that
\begin{align*}
\overset{\cdot}{W}& =\overline{V}W=\frac{1}{2}\left(  fD_{r}f^{-1}
(I+C_{H})+f^{-1}D_{r}f(I-C_{H})\right)  W \\*[2ex] = &
\frac{1}{2}\left(  fD\mathcal{A}\left[  \frac{\mathbf{w}}{f}\right]
+f^{-1}D\operatorname*{rot}\left(  \mathbf{B}\left[  f\mathbf{w}\right]
\right)  \right)  =\frac{1}{2}\left(  \mathbf{w+w}\right)  =\mathbf{w}.
\end{align*}
\end{proof}

Another important corollary of the preceding calculations is the following
generalization of theorem \ref{ThConjHarm}.

\begin{theorem}
Let $W_{0}$ be a scalar solution of (\ref{Schro3d}) with $q=\frac{\triangle
f}{f}$ in $\Omega$. Then the vector function $\mathbf{W}$ such that
$\mathbf{\Phi}=f\mathbf{W}$ be a solution of (\ref{eqPhi}) and (\ref{divPhi})
and $W=W_{0}+\mathbf{W}$ be a solution of (\ref{mainVekuaquat}), is
constructed according to the formula
\begin{equation}
\mathbf{W}=-f^{-1}\left\{  \text{\emph{rot}}\left(  \mathbf{B}\left[
f^{2}\ \nabla\left(  f^{-1}W_{0}\right)  \right]  \right)  +\nabla h\right\}
. \label{Wvect}
\end{equation}

Given a solution $\mathbf{\Phi}$ of (\ref{eqPhi}) and (\ref{divPhi}), for
$\mathbf{W=}\frac{1}{f}\mathbf{\Phi}$ the corresponding solution $W_{0}$ of
(\ref{Schro3d}) such that $W=W_{0}+\mathbf{W}$ be a solution of
(\ref{mainVekuaquat}), is constructed as follows
\begin{equation}
W_{0}=-f\mathcal{A}\left[  f^{-2}\operatorname*{rot}(f\mathbf{W})\right]  .
\label{W0}
\end{equation}

\end{theorem}

\begin{proof}
Let $W_{0}$ be a scalar solution of (\ref{Schro3d}) with $q=\frac{\triangle
f}{f}$. Then from (\ref{Dphi0}) we have that $\overset{\cdot}{W}
=2f\nabla\left(  f^{-1}W_{0}\right)  $ and due to (\ref{Phi}), we obtain
\[
\mathbf{\Phi}=-\operatorname*{rot}\left(  \mathbf{B}\left[  f^{2}
\ \nabla\left(  f^{-1}W_{0}\right)  \right]  \right)  +\nabla h,
\]
and as a consequence (\ref{Wvect}) holds.

Now let $\mathbf{\Phi}$ be a solution of (\ref{eqPhi}) and (\ref{divPhi}).
According to (\ref{DPhi}) we have $\overset{\cdot}{W}=-2f^{-1}D\mathbf{\Phi
=}-2f^{-1}\operatorname*{rot}\left(  f\mathbf{W}\right)  $. Now from
(\ref{phi0}) we obtain (\ref{W0}).
\end{proof}

\section{Examples and applications}

In the present section we analize the possibility to construct solutions for
the equation (\ref{D+M}) under certain restrictions on the function $f$ and
apply them for obtaining solutions of the Schr\"{o}dinger equation
(\ref{Schro3d}).

\begin{proposition}
\label{PropParallel}Let $f$ be a nonvanishing $C^{1}$ scalar function defined
in $\Omega$ and there exist a harmonic function $\rho$ such that
\begin{equation}
\nabla f\times\nabla\rho  =0\text{\quad in }\Omega
.\label{parallel}
\end{equation}
Then the purely vectorial function
\begin{equation}
\mathbf{F}=\frac{1}{f}D\rho\label{F}
\end{equation}
is a solution of (\ref{D+M}) and $\mathbf{G}=fD\rho$ is a solution of the
equation
\begin{equation}
\left(  D-M^{\frac{Df}{f}}\right)  \mathbf{G}=0.\label{D-M}
\end{equation}

\end{proposition}

\begin{proof}
Consider $D\mathbf{F}=-\frac{1}{f^{2}}Df\,D\rho=-\frac{1}{f}D\rho\,\frac
{Df}{f}=-\mathbf{F}\frac{Df}{f}$. In a similar way one can check the validity
of (\ref{D-M}).
\end{proof}

Let us notice that (\ref{parallel}) is fulfilled when $f$ depends on $\rho$
and this is of a considerable importance because harmonic functions are behind
all physically interesting orthogonal coordinate systems. For example,
consider the spherical coordinate $r=\sqrt{x^{2}+y^{2}+z^{2}}$ and let
$f=f(r)$. The function $r$ itself is not harmonic, however the function
$\rho=1/r$ is harmonic everywhere except the origin. Thus, when $f$ is a
function of $r$ and hence a function of $\rho$, and the domain $\Omega$ does
not contain the origin the function (\ref{F}) is a solution of (\ref{D+M}).

\begin{proposition}
\label{PropOrthogonal}Let $f$ be a nonvanishing $C^{1}$ scalar function
defined in $\Omega$ and there exist a harmonic function $\rho$ such that
\[
\left\langle \nabla f,\nabla\rho\right\rangle =0\text{\quad in }\Omega.
\]
Then the purely vectorial function
\[
\mathbf{F}=fD\rho
\]
is a solution of (\ref{D+M}) and $\mathbf{G}=\frac{1}{f}D\rho$ is a solution
of (\ref{D-M}).
\end{proposition}

\begin{proof}
Consider $D\mathbf{F}=Df\,D\rho=-fD\rho\,\frac{Df}{f}=-\mathbf{F}\frac{Df}{f}
$. In a similar way one can check the validity of (\ref{D-M}).
\end{proof}

These two propositions give us the possibility to obtain three independent
solutions of equation (\ref{D+M}) in a simple way when $f$ is a function of a
harmonic function and this harmonic function possesses two different
orthogonal harmonic functions. As we show in the next example this situation
is of practical importance. Let us notice that in general for a given harmonic
function in three dimensions there not necessarilly exists a nonconstant
orthogonal harmonic function (a counterexample is given in \cite{Elkins}).

\begin{example}
Consider the cylindrical coordinate system $r=\sqrt{x^{2}+y^{2}}$,
$\theta=\arctan(y/x)$ and $z$. Suppose that $f$ depends only on $r$. Then the
required harmonic function $\rho$ can be chosen as $\rho=\log r$, and
obviously the functions $\theta$ and $z$ are harmonic and orthogonal to $\rho
$. Thus $\mathbf{F}_{1}=\frac{1}{f(r)}D\rho=-\frac{1}{f(r)}\left(  \frac
{x}{r^{3}}\mathbf{e}_{1}+\frac{y}{r^{3}}\mathbf{e}_{2}\right)  $,
$\mathbf{F}_{2}=f(r)D\theta=-f(r)\left(  -\frac{y}{r^{2}}\mathbf{e}_{1}
+\frac{x}{r^{2}}\mathbf{e}_{2}\right)  $ and $\mathbf{F}_{3}=f(r)\mathbf{e}
_{3}$ are solutions of (\ref{D+M}) independent in the sense that any function
$\mathbf{w}$ defined in a domain not containing the axis $r=0$ and having
purely quaternionic values can be represented in the form $\mathbf{w}
=\sum_{k=1}^{3}\varphi_{k}\mathbf{F}_{k}$ where $\varphi_{k}$ are scalar
functions. Consequently $\mathbf{F}_{1}$, $\mathbf{F}_{2}$ and $\mathbf{F}
_{3}$ form a generating triplet of equation (\ref{D+M}) (see more on this
concept in \cite{KR2010}) and $\mathbf{w}$ is a solution of (\ref{D+M}) if and
only if the scalar functions $\varphi_{k}$ satisfy the equation
\begin{equation}
\sum_{k=1}^{3}D\varphi_{k}\,\mathbf{F}_{k}=0\text{\quad in }\Omega
.\label{Vekua3Dseckind}
\end{equation}
In general we note that if $f$ is a function of a harmonic function depending
on two Cartesian variables then we always can obtain a generating triplet for
equation (\ref{D+M}) and hence reduce the equation to the form
(\ref{Vekua3Dseckind}). Indeed, for such harmonic function we always have two
orthogonal harmonic functions: its conjugate harmonic depending on two
Cartesian variables as well, and the third Cartesian coordinate.
\end{example}

Assume that $f$ is a function of a harmonic function $\rho$. Then according to
proposition \ref{PropParallel}, $\mathbf{F}=\frac{1}{f}D\rho$ is a solution of
(\ref{D+M}). Its antiderivative (\ref{Antider}) gives us a solution of
(\ref{VekuaMain3DExplicit}) and according to theorem
\ref{ThSolutionsOfVekuaMain3D} the scalar part of the obtained solution is a
solution of (\ref{Schro3d}). Thus we obtain that $\psi=f\mathcal{A}\left[
\frac{\mathbf{F}}{f}\right]  =f\mathcal{A}\left[  \frac{D\rho}{f^{2}}\right]
$ is necessarily a solution of (\ref{Schro3d}), and hence the results of this
section allow us to construct new solutions of the Schr\"{o}dinger equation (\ref{Schro3d}) from known
solutions possessing certain symmetries.

\section{Conclusions}

For biquaternionic Vekua type equations arising from the factorization of
linear second order elliptic operators we introduced and studied the concepts
of the derivative and of the antiderivative in the sense of Bers. These
concepts represent new relations between first order elliptic systems and
allow us to study solutions of the second order elliptic equations in relation
to components of the biquaternion valued solutions of biquaternionic Vekua
equations. We have shown that the revealed relations are useful for obtaining
new exact solutions of the Schr\"{o}dinger type equations and at the same time
they imply natural open questions important for the general theory of linear
second order elliptic equations. Namely, we showed how in a quite general and
important for practical applications situation one can obtain a generating
triplet for the biquaternionic Vekua equation which can be considered as a
successor of the main Vekua equation (\ref{VekuaMain3DExplicit}). In the
two-dimensional case this was sufficient to obtain a whole generating sequence
related to the Vekua equation and as a consequence a complete system of
solutions called formal powers (see \cite{APFT}) for the Vekua equation and
for the related second order elliptic equation. At present it is not clear how
to generalize those results onto the spatial situation.

\subsection*{Acknowledgments}
V.K. acknowledges the support of CONACYT of Mexico via the research project 50424.
The research of S.T. is partly supported by grant from NSERC of Canada.


TITLE: Not sure how to solve Bayesian parameter learning problem
QUESTION [2 upvotes]: I could use some help solving a problem about a Dirichlet prior.
We have a multinomial distribution over an alphabet of 27 symbols parameterized by $\mathbf{\theta}=(\theta_1, ..., \theta_{27})$. We have collected a dataset $D = \{x[1],\ldots,x[2000]\}$ consisting of $2000$ symbols, among which "e" appears $260$ times.  We use a Dirichlet prior over $\theta$ with parameters $\mathbf{\alpha}= (\alpha_1, ..., \alpha_{27})$,  where each $\alpha_i = 10$.
 What is the predictive probability that letter "e" occurs with this prior? (i.e., what is $P(X[2001] = $ "$e$"$ \mid D)$?)
So my understanding is that we are to find  $P(X_{2001} | X_1, X_2, ... X_{2000})$. By the given information, and since this is a Bayesian probability question, it appears that we must  apply the chain rule of conditional probability here. If we also observe the conditional independence of $X_1, X_2, ... X_{2001}$ given $\theta$, we have
$$P(X_{2001} | X_1, ... X_{2000}, \theta)P(\theta|X_1, X_{2000})= P(X_{2001}  |\theta)P(\theta|X_1, X_{2000})$$
and from law of total probability,
$$ =\int_{\theta} P(X_{2001} |\theta)P(\theta|X_1, ... X_{2000}) \mathrm d\theta  $$
Now to solve the integral, I am left with some questions. What is $P(X_{2001}|\theta)$?  Is this simply $\theta$? Can this probability be derived from the multinomial of $X$ given $\alpha$?
How do we find the posterior $P(\theta|X_1, ... X_{2000})$? I was thinking that since the Dirichlet (Beta) is a conjugate prior to the multinomial (binomial), that this posterior might be
$$P(\theta|X_1, ... X_{2000}) = \frac{\Gamma(\alpha_5+  260 + \sum_{i=1}^{26} \alpha_i + \frac{2000-260}{26})}{\Gamma(\alpha_5 + 260)\Pi_{i=1}^{26}\Gamma(\alpha_i + \frac{2000-260}{26})} \theta_{5}^{10+260-1}\Pi_{i=1}^{26}\theta_{i}^{1000/13 - 1}$$
Where the subscript $5$ denotes the index for "e" in the parameter vectors. $\alpha_i$ is the value of the $i$th value in the vector $\alpha$ after the $2000$ observations (inclusive of the starting prior parameter values of $\alpha_i=10$). Next, substituting the given information that "e" was observed $260$ of the $2000$ times gives me
$$P(\theta|X_1, ... X_{2000}) = \frac{\Gamma(260)\Gamma(1740)}{\Gamma(260)\Pi_{i=1}^{26}\Gamma(\alpha_i))} \theta_{5}^{260}\Pi_{i=1}^{1740}\theta_{i}^{\alpha_i}$$
I am very unsure of my work so far, and I am not sure how to finish solving this. The last expression is in terms of the vector $\theta$, and I would not know how to integrate such an expression.
I was wondering if it is possible to simplify this problem by modeling the distribution over $\mathbf{\theta}$ as a binomial distribution instead of a Dirichlet. It appears that is really only a binary event here: either "e" is observed or some other symbol was observed.

REPLY [2 votes]: What happens is that you start from the Dirichlet distribution as a prior, i.e.
$$p(\theta)=\frac{\Gamma(\sum_{i=1}^{27} \alpha_i)}{\Pi_{i=1}^{27}\Gamma(\alpha_i)} \Pi_{i=1}^{27}\theta_i^{\alpha_i}$$
Then, you obtain some data which have a certain probability under the hypothesis that the $\alpha_i$ have certain values.
$$P(X_1,\ldots,X_{2000}|\theta)={2000 \choose {N_1,\ldots,N_{27}}} \Pi_{i=1}^{27}{\theta_i}^{N_i}$$
In which an $N_i=\sum_k I(X_k=i)$ or in words $N_i$ counts how many times the $i$th symbol appears. From this you now compute the posterior as
$$P(\theta|X_1,\ldots,X_{2000})=\frac{P(X_1,\ldots,X_{2000}|\theta)p(\theta)}{\int P(X_1,\ldots,X_{2000}|\theta)p(\theta) d\theta}$$
However, you only have partial information about the letter "e". Let's say that letter "e" is the 5th symbol, then the probability of it appearing 260 times under your hypothesis is
$$P(N_5=260|\theta)={2000 \choose 260} {\theta_5}^{260}(1-\theta_5)^{2000-260}$$
and applying the Bayesian updating rule we should write
$$P(\theta|N_5=260)=\frac{P(N_5=260|\theta)p(\theta)}{\int P(N_5=260|\theta)p(\theta) d\theta}$$
Working out the integral
$$\int P(N_5=260|\theta)p(\theta) d\theta = {2000 \choose 260} \frac{\Gamma(\sum_{i=1}^{27} \alpha_i)}{\Pi_{i=1}^{27}\Gamma(\alpha_i)} \int {\theta_5}^{260}(1-\theta_5)^{2000-260}  \prod_{i=1}^{27}\theta_i^{\alpha_i} \Pi_{i=1}^{27}d\theta_i$$
The factor in front is irrelevant since it will cancel with the one from the numerator in our end formula, let's concentrate on
$$\int {\theta_5}^{260}(1-\theta_5)^{2000-260}  \prod_{i=1}^{27}\theta_i^{\alpha_i} \Pi_{i=1}^{27}d\theta_i = \int {\theta_5}^{\alpha_5+260}(\sum_{i\neq 5}\theta_i)^{2000-260}  \prod_{i\neq 5}\theta_i^{\alpha_i} \Pi_{i=1}^{27}d\theta_i$$
We work out the $(\sum_{i\neq 5}\theta_i)^{2000-260}$ with the multinomial theorem
$$(\sum_{i\neq 5}\theta_i)^{2000-260}=\sum_{k_i}{{2000-260} \choose {k_1,\ldots,\hat{k}_5\ldots,k_{27}}}\prod_{i\neq 5}\theta_i^{k_i}$$
which gives us for the integral
$$\int {\theta_5}^{\alpha_5+260}(\sum_{i\neq 5}\theta_i)^{2000-260}  \prod_{i\neq 5}\theta_i^{\alpha_i} \Pi_{i=1}^{27}d\theta_i = \sum_{k_i}{{2000-260} \choose {k_1,\ldots,\hat{k}_5\ldots,k_{27}}}\frac{\Gamma(\alpha_5+260)\prod_{i\neq 5}\Gamma(\alpha_i+k_i)}{\Gamma(\sum_i \alpha_i + 2000)}$$


TITLE: Retraction of a Riemannian manifold with boundary to its cut locus
QUESTION [11 upvotes]: This question is edited following the comment of Joseph. He pointed out that the main object of the first version of this question is the cut locus.
Recall that the cut locus of a set $S$ in a geodesic space $X$
is the closure of the set of all points $p \in X$ that have two or more distinct shortest paths in $X$ from $S$ to $p$.
http://en.wikipedia.org/wiki/Cut_locus
A simple lemma shows that, for a disk $D^2$ with a Riemannian metric and piecewise smooth generic boundary, the cut locus of $D^2$ with respect to its boundary is a tree.
A picture of such tree can be found on page 542, figure 17 of the article of Thurston "Shapes of polyhedra". The tree is white.
http://arxiv.org/PS_cache/math/pdf/9801/9801088v2.pdf
For an ellipse on the 2-plane, the tree is the segment that joins its focal points.
More generically for a Riemannian manifold $M^n$ with boundary, the cut locus of $\partial M$ should be a deformation retract of $M$.  (I guess it is a $CW$ complex of dimension less than $n$.) To prove this lemma, notice that $M^n\setminus\operatorname{cut-locus}(\partial M^n)$ is canonically foliated by geodesic segments that join $X$ with $\partial M$.
I wonder if this lemma has a name or maybe is contained in some textbook on Riemannian geometry?

REPLY [6 votes]: The 'simple lemma' certainly isn't always true.  Let's consider just open disks in the Euclidean plane with a flat metric. Fremlin (Proc. LMS, 1997) calls the set of points without a unique nearest boundary point the "skeleton".  He gives an example where the skeleton is somewhere dense (that is, its closure, which you call the "cut locus", has nonempty interior).  The skeleton will always be an R-tree.


TITLE: Is each space filling curve everywhere self-intersecting?
QUESTION [3 upvotes]: Consider a continuous surjection $f:[0,1]\to [0,1]\times[0,1]$. Is
$$\{x:\exists(t_1\not=t_2) f(t_1)=f(t_2)=x\}=[0,1]\times[0,1]?$$

REPLY [6 votes]: No. Consider for instance the Hilbert space-filling curve, on which Brian Hayes wrote a nice popular article recently (HTML, PDF).
This mapping from $[0, 1]$ to $[0, 1] \times [0, 1]$ can be viewed as a mapping from digit sequences $0.d_1d_2d_3\dots$ in base-$4$ (where $0 \le d_i < 4$) to $[0, 1] \times [0, 1]$. The first digit tells us which quadrant a point lies in, the second digit tells us which sub-quadrant of that it lies in, and so on. (See the second image on page 4.)
Conversely, given a point in $[0, 1] \times [0, 1]$, we can invert the curve -- find all $t$ in $[0, 1]$ that are mapped to this point -- by noting down which quadrant it lies in, then which sub-quadrant of that it lies in, and so on, recursively.
The only points that are the image of multiple $t$ are those that, at some granularity, lie in multiple quadrants. These are precisely those points $(a, b)$ such that either $a$ or $b$ can be written as $\dfrac{m}{2^k}$ for some integers $k \ge 1$ and $1 \le m < 2^k$. Every other point is the image of exactly one $t$.
So for the Hilbert curve at least, the set of points at which the curve is "self-intersecting" in your sense is, far from being every point, actually of measure $0$.


TITLE: Why does $|e^{2iz}|\leq |e^{2i(x+iy)}|\leq |e^{2ix}||e^{-2y}|\leq 1$ for $z$ in the upper half plane?
QUESTION [0 upvotes]: Why does $|e^{2iz}|\leq |e^{2i(x+iy)}|\leq |e^{2ix}||e^{-2y}|\leq 1$ for $z$ in the upper half plane?
Is it just because it is a complex number with modulus $1$ and all you are doing by changing $z$ is just rotating it around but not actually changing it's modulus?
If that is the case then why do the steps need to be written out as above surely $|e^{2iz}|=1$ would make more sense to write?
Any assistance?
Context: Using a semicircle contour to evaluate a real integral using complex integral.
http://imgur.com/OVtil3l

REPLY [0 votes]: You have written it slightly differently than the image. You have that
$$|e^{2iz}| = |e^{2ix}||e^{-2y}| = 1\cdot |e^{-2y}| = e^{-2y}. $$
So the magnitude depends on $y$, i.e. the imaginary part of $z$.
Futhermore, we have that $e^{-2y}\leq 1$ for all $y\geq 0$, because $e^{-2y}$ is strictly decreasing as $y$ increases, and $e^{-2\cdot 0} = 1$.


TITLE: Identifying $\langle a,b\mid a^2b^2\rangle$.
QUESTION [3 upvotes]: The Question:

What group is $G=\langle a,b\mid a^2b^2\rangle$?

Thoughts:
I found that the presentation maps onto $\langle a, b\mid a^2, b^2, 1\cdot 1\rangle\cong \mathbb{Z}_2\ast\mathbb{Z}_2$, which is reassuring since GAP says $G$ is infinite.
Extra Context:
I'm not sure if that's enough context so here's a Q&A:

What are you studying?

A PhD in combinatorial group theory, first year.

What text is this drawn from, if any?  If not, how did the question arise?

None. The group arose in my research. Without giving too much away, the group is cyclically presented.

What kind of approaches (to similar problems) are you familiar with?

Here's a list of related questions of mine:
Identifying $\langle \{x_h\mid h\in H\}\mid \{x_{h'}x_{h'a}x_{h'b}^{-1}\mid h'\in H\}\rangle$ if $H=\langle a,b\mid a^2, b^4, ab=ba\rangle$.
Identifying $\langle r,s\mid srs^{-2}r, r^{-1}srsr^{-1}\rangle$.
Identifying $\langle x_e, x_a, x_b, x_{ab}\mid x_ex_a=x_{ab}, x_ax_e=x_b, x_bx_{ab}=x_a, x_{ab}x_b=x_e\rangle.$
Identifying $\langle x_1, \dots, x_6\mid x_1x_2=x_3, x_2x_3=x_1, x_3x_1=x_2, x_4x_5=x_6, x_5x_6=x_4, x_6x_4=x_5\rangle.$
There's more but I think that's enough.

What kind of answer are you looking for?  Basic approach, hint, explanation, something else?

A simple identification would be great. A detailed explanation of why it is what it is would be better, although (strong) hints are preferred.

Is this question something you think should be able to answer?  Why or why not?

Yes. I've been working closely with presentations in a research context for months now. It looks like Google should have the answer.
Please help :)

REPLY [4 votes]: For a topological/geometric point of view, this group is isomorphic to the fundamental group of the Klein bottle.
Here's how to see it. Start by rewriting the presentation
$$\langle a,b \mid a^2b^2 = \text{Id}\rangle = \langle a,b \mid a^2 = b^{-2} \rangle = \langle a,c \mid a^2 = c^2 \rangle \quad\text{($c=b^{-1}$)}
$$
By Van Kampen's Theorem, we can construct a topological space with this fundamental group. Start with two circles $C_a$, $C_c$ whose fundamental groups are infinite cyclic with generators represented by closed paths $\gamma_a$ going exactly one time around $C_a$ and $\gamma_c$ going one time around $C_c$. Now attach $S^1 \times [0,1]$ to these two circles: $S^1 \times 0$ is attached to the path $\gamma_c^2$ that goes two times around $C_a$; and $S_1 \times 1$ is attached to the path $\gamma_b^2$ that goes two times around $C_c$. Let $X$ be the resulting quotient space. This space $X$ is a compact, connected, non-orientable 2-manifold of Euler characteristic $0$, and so it is homeomorphic to the Klein bottle. Thus, by Van Kampen's Theorem, your group is isomorphic to the fundamental group of the Klein bottle.
And once you know that this is the Klein bottle group, it also follows that it is one of the 17 famous wallpaper groups, also known as planar crystallographic groups, namely those groups which can be represented as discrete, cocompact subgroups of isometries of the Euclidean plane. To a crystallographer, this group is known as "pg", and can be represented as generated by two isometries of the $xy$-plane, as shown in this picture: $g_x$ which is a glide reflection along the $x$ axis (the "g" in the "pg" notation refers to a glide reflector); and $t_y$ which is a translation along the $y$ axis. To connect these generators with your two original generators, you can take $a=g_x$ and $b=t_y g_x$.


TITLE: Heat equation with discontinuous sink and zero flux boundary conditions
QUESTION [4 upvotes]: I am trying to solve the following problem:
$\frac{\partial p}{\partial t}=D\frac{\partial^2 p}{\partial r^2}-\lambda p,$
on the domain $r\in[0,r_w]$ and for times $t\in [0,\infty)$, where
$ {\displaystyle \lambda={\begin{cases}\lambda_s,& \text{for } 0<r\leq r_s,
\\ \lambda_w,& \text{for } r_s<r\leq r_w.\end{cases}}}$
The initial condition is a dirac delta function at the right hand end of the domain
$p(r,0)=\delta(r-r_w),$
and the boundary conditions are zero flux at either end of the domain:
$\frac{\partial p}{\partial r}|_{r=0}=0$ and $\frac{\partial p}{\partial r}|_{r=r_w}=0$.
At the intermediate boundary I would like continuity and smoothness of the solution:
$p_I(r_s)=p_O(r_s)$,
and
$\frac{\partial p_I}{\partial r}|_{r=r_s}=\frac{\partial p_O}{\partial r}|_{r=r_s}$,
where $p_I$ and $p_O$ are the inner and outer solutions respectively.
If there is a reason why these can't be satisfied, I would be interested to hear why.
I have tried fourier series solutions for the two distinct parts of the domain but couldn't get the matching correct. Any help would be appreciated.

REPLY [5 votes]: Part II
Proceeding now with the actual problem.  We define two functions over the two domains,
${{u}_{-}}\left( x,t \right),0\le x\le {{x}_{s}}$and ${{u}_{+}}\left( x,t \right),{{x}_{s}}<x\le 1$.  Then proceeding as we did before we find
$${{\hat{u}}_{-t}}\left( -ik,t \right)+\left( {{k}^{2}}+{{\lambda }_{-}} \right){{\hat{u}}_{-}}\left( -ik,t \right)=\left( {{u}_{-x}}\left( {{x}_{s}},t \right)+ik{{u}_{-}}\left( {{x}_{s}},t \right) \right){{e}^{-ik{{x}_{s}}}}-\left( {{u}_{-x}}\left( 0,t \right)+ik{{u}_{-}}\left( 0,t \right) \right)$$
$${{\hat{u}}_{+t}}\left( -ik,t \right)+\left( {{k}^{2}}+{{\lambda }_{+}} \right){{\hat{u}}_{+}}\left( -ik,t \right)=\left( {{u}_{+x}}\left( 1,t \right)+ik{{u}_{+}}\left( 1,t \right) \right){{e}^{-ik}}-\left( {{u}_{+x}}\left( {{x}_{s}},t \right)+ik{{u}_{+}}\left( {{x}_{s}},t \right) \right){{e}^{-ik{{x}_{s}}}}$$
Writing the LHS as a derivative, then integrating we have
$${{\hat{u}}_{-}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-{{\hat{u}}_{-}}\left( -ik,0 \right)=\int\limits_{0}^{t}{\left( {{u}_{-x}}\left( {{x}_{s}},y \right)+ik{{u}_{-}}\left( {{x}_{s}},y \right) \right){{e}^{-ik{{x}_{s}}+\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}-\left( {{u}_{-x}}\left( 0,y \right)+ik{{u}_{-}}\left( 0,y \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}dy}$$
$${{\hat{u}}_{+}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}-{{\hat{u}}_{+}}\left( -ik,0 \right)=\int\limits_{0}^{t}{\left( {{u}_{+x}}\left( 1,y \right)+ik{{u}_{+}}\left( 1,y \right) \right){{e}^{-ik+\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}-\left( {{u}_{+x}}\left( {{x}_{s}},y \right)+ik{{u}_{+}}\left( {{x}_{s}},y \right) \right){{e}^{-ik{{x}_{s}}+\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}$$
As done previously we define the following functions by writing the above equations as
$${{\hat{u}}_{-}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-{{\hat{u}}_{-}}\left( -ik,0 \right)=\left( {{{\tilde{j}}}_{-1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{-0}}\left( {{k}^{2}} \right) \right){{e}^{-ik{{x}_{s}}}}-\left( {{{\tilde{g}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{g}}}_{0}}\left( {{k}^{2}} \right) \right)$$
$${{\hat{u}}_{+}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}-{{\hat{u}}_{+}}\left( -ik,0 \right)=\left( {{{\tilde{h}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{h}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{-ik}}-\left( {{{\tilde{j}}}_{+1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{+0}}\left( {{k}^{2}} \right) \right){{e}^{-ik{{x}_{s}}}}$$
The j-functions represent the boundary value data for the intermediate layer.
Inverting the transform and deforming the integrals to a contour off the real line (the boundaries/regions are the same as before) we have
$$\begin{align}
  & {{u}_{-}}\left( x,t \right)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{{{{\hat{u}}}_{-}}\left( -ik,0 \right){{e}^{ikx-\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}dk} \\
 & -\frac{1}{2\pi }\int\limits_{\partial {{D}^{+}}}^{{}}{\left( {{{\tilde{g}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{g}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{ikx-\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}dk}-\frac{1}{2\pi }\int\limits_{\partial {{D}^{-}}}^{{}}{\left( {{{\tilde{j}}}_{-1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{-0}}\left( {{k}^{2}} \right) \right){{e}^{ikx-ik{{x}_{s}}-\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}dk} \\
\end{align}$$
$$\begin{align}
  & {{u}_{+}}\left( x,t \right)=\frac{1}{2\pi }\int\limits_{-\infty }^{\infty }{{{{\hat{u}}}_{+}}\left( -ik,0 \right){{e}^{ikx-\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}} \\
 & -\frac{1}{2\pi }\int\limits_{\partial {{D}^{+}}}^{{}}{\left( {{{\tilde{j}}}_{+1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{+0}}\left( {{k}^{2}} \right) \right){{e}^{ikx-ik{{x}_{s}}-\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}dk}+\frac{1}{2\pi }\int\limits_{\partial {{D}^{-}}}^{{}}{\left( {{{\tilde{h}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{h}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{ikx-ik-\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}dk} \\
\end{align}$$
At the moment we have eight transformed boundary value points and only two equations. The symmetry in the unknown transformed boundary data allows us to map $k\to -k$ , which then yields four equations.  Substituting known boundary conditions $${{\tilde{g}}_{1}}\left( {{k}^{2}} \right)={{\tilde{h}}_{1}}\left( {{k}^{2}} \right)=0$$eliminates two unknowns and so we have 6 unknowns and four equations.  Two more conditions are required.
Note from the differential equation for either function we have
$\frac{\partial {{u}_{\pm }}}{\partial t}=\frac{{{\partial }^{2}}{{u}_{\pm }}}{\partial {{x}^{2}}}-{{\lambda }_{\pm }}{{u}_{\pm }}\Rightarrow \frac{\partial }{\partial t}\int\limits_{{{x}_{s}}-\varepsilon }^{{{x}_{s}}+\varepsilon }{{{u}_{\pm }}dx}={{\left. \frac{\partial {{u}_{+}}}{\partial x} \right|}_{{{x}_{s}}+\varepsilon }}-{{\left. \frac{\partial {{u}_{-}}}{\partial x} \right|}_{{{x}_{s}}-\varepsilon }}-{{\lambda }_{\pm }}\int\limits_{{{x}_{s}}-\varepsilon }^{{{x}_{s}}+\varepsilon }{{{u}_{\pm }}dx}$
If the function is ‘well-behaved’ on the boundary then the integrals go to zero as epsilon goes to zero and we have
${{\left. \frac{\partial {{u}_{+}}}{\partial x} \right|}_{{{x}_{s}}}}={{\left. \frac{\partial {{u}_{-}}}{\partial x} \right|}_{{{x}_{s}}}}$
Note that for example we have therefore
$${{\tilde{j}}_{+1}}\left( {{k}^{2}} \right)=\int\limits_{0}^{t}{{{u}_{+x}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}=\int\limits_{0}^{t}{{{u}_{-x}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}\ne {{\tilde{j}}_{-1}}\left( {{k}^{2}} \right),{{\tilde{j}}_{+0}}\left( {{k}^{2}} \right),{{\tilde{j}}_{-0}}\left( {{k}^{2}} \right)$$
And so this in no way allows us to eliminate an unknown unless ${{\lambda }_{+}}={{\lambda }_{-}}$, in which case ${{\tilde{j}}_{+1}}\left( {{k}^{2}} \right)={{\tilde{j}}_{-1}}\left( {{k}^{2}} \right)$.  But obviously we don’t want this.   A similar situation exists if we insist that the ‘temperatures’ on the boundary are the same, i.e. ${{u}_{+}}\left( {{x}_{s}},y \right)={{u}_{-}}\left( {{x}_{s}},y \right)$, in that this conditions won’t allow for the elimination of  ${{\tilde{j}}_{+0}}\left( {{k}^{2}} \right)$ or ${{\tilde{j}}_{-0}}\left( {{k}^{2}} \right)$ .
Note however that if we insist ${{u}_{+x}}\left( {{x}_{s}},t \right)={{e}^{\left( {{\lambda }_{-}}-{{\lambda }_{+}} \right)t}}{{u}_{-x}}\left( {{x}_{s}},t \right)$, and ${{u}_{+}}\left( {{x}_{s}},t \right)={{e}^{\left( {{\lambda }_{-}}-{{\lambda }_{+}} \right)t}}{{u}_{-}}\left( {{x}_{s}},t \right)$ then we can eliminate two more unknowns, for note:
$${{\tilde{j}}_{+1}}\left( {{k}^{2}} \right)=\int\limits_{0}^{t}{{{u}_{+x}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}=\int\limits_{0}^{t}{{{u}_{-x}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}dy}={{\tilde{j}}_{-1}}\left( {{k}^{2}} \right)$$
$${{\tilde{j}}_{+0}}\left( {{k}^{2}} \right)=\int\limits_{0}^{t}{{{u}_{+}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}=\int\limits_{0}^{t}{{{u}_{-}}\left( {{x}_{s}},y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}dy}={{\tilde{j}}_{-0}}\left( {{k}^{2}} \right)$$
But as we will see, the solution this generates has problems, and so therefore the assumed intermediate boundary conditions appear invalid with respect to the system.  However for now let us proceed with these assumptions. Doing so we have four equations and four unknowns.
$${{\hat{u}}_{-}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-{{\hat{u}}_{-}}\left( -ik,0 \right)=\left( {{{\tilde{j}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{-ik{{x}_{s}}}}-ik{{\tilde{g}}_{0}}\left( {{k}^{2}} \right)$$
$${{\hat{u}}_{+}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}-{{\hat{u}}_{+}}\left( -ik,0 \right)=ik{{\tilde{h}}_{0}}\left( {{k}^{2}} \right){{e}^{-ik}}-\left( {{{\tilde{j}}}_{1}}\left( {{k}^{2}} \right)+ik{{{\tilde{j}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{-ik{{x}_{s}}}}$$
$${{\hat{u}}_{-}}\left( ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-{{\hat{u}}_{-}}\left( ik,0 \right)=\left( {{{\tilde{j}}}_{1}}\left( {{k}^{2}} \right)-ik{{{\tilde{j}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{ik{{x}_{s}}}}+ik{{\tilde{g}}_{0}}\left( {{k}^{2}} \right)$$
$${{\hat{u}}_{+}}\left( ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}-{{\hat{u}}_{+}}\left( ik,0 \right)=-ik{{\tilde{h}}_{0}}\left( {{k}^{2}} \right){{e}^{ik}}-\left( {{{\tilde{j}}}_{1}}\left( {{k}^{2}} \right)-ik{{{\tilde{j}}}_{0}}\left( {{k}^{2}} \right) \right){{e}^{ik{{x}_{s}}}}$$
 Solving this system of equations we find
$${{\tilde{g}}_{0}}\left( {{k}^{2}} \right)=\frac{i}{k\left( {{e}^{2ik}}-1 \right)}\left( \begin{align}
  & -{{e}^{2ik}}\left( {{{\hat{u}}}_{-}}\left( -ik,0 \right)+{{{\hat{u}}}_{+}}\left( -ik,0 \right) \right)-\left( {{{\hat{u}}}_{-}}\left( ik,0 \right)+{{{\hat{u}}}_{+}}\left( ik,0 \right) \right) \\
 & +\left( {{e}^{2ik}}{{{\hat{u}}}_{-}}\left( -ik,t \right)+{{{\hat{u}}}_{-}}\left( ik,t \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}+\left( {{e}^{2ik}}{{{\hat{u}}}_{+}}\left( -ik,t \right)+{{{\hat{u}}}_{+}}\left( ik,t \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}} \\
\end{align} \right)$$
$${{\tilde{h}}_{0}}\left( {{k}^{2}} \right)=\frac{i{{e}^{ik}}}{k\left( {{e}^{2ik}}-1 \right)}\left( \begin{align}
  & -\left( {{{\hat{u}}}_{-}}\left( -ik,0 \right)+{{{\hat{u}}}_{-}}\left( ik,0 \right)+{{{\hat{u}}}_{+}}\left( -ik,0 \right)+{{{\hat{u}}}_{+}}\left( ik,0 \right) \right) \\
 & +\left( {{{\hat{u}}}_{-}}\left( -ik,t \right)+{{{\hat{u}}}_{-}}\left( ik,t \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}+\left( {{{\hat{u}}}_{+}}\left( -ik,t \right)+{{{\hat{u}}}_{+}}\left( ik,t \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}} \\
\end{align} \right)$$
$${{\tilde{j}}_{0}}\left( {{k}^{2}} \right)=-\frac{i{{e}^{-ik{{x}_{s}}}}}{2k\left( {{e}^{2ik}}-1 \right)}\left( \begin{align}
  & \left( {{{\hat{u}}}_{-}}\left( -ik,0 \right)+{{{\hat{u}}}_{-}}\left( ik,0 \right) \right)\left( {{e}^{2ik}}+{{e}^{2ik{{x}_{s}}}} \right)+\left( {{e}^{2ik}}{{{\hat{u}}}_{+}}\left( -ik,0 \right)+{{{\hat{u}}}_{+}}\left( ik,0 \right) \right)\left( 1+{{e}^{2ik{{x}_{s}}}} \right) \\
 & -\left( {{{\hat{u}}}_{-}}\left( -ik,t \right)+{{{\hat{u}}}_{-}}\left( ik,t \right) \right)\left( {{e}^{2ik}}+{{e}^{2ik{{x}_{s}}}} \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-\left( {{{\hat{u}}}_{+}}\left( -ik,t \right){{e}^{2ik}}+{{{\hat{u}}}_{+}}\left( ik,t \right) \right)\left( 1+{{e}^{2ik{{x}_{s}}}} \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}} \\
\end{align} \right)$$
$${{\tilde{j}}_{1}}\left( {{k}^{2}} \right)=\frac{{{e}^{-ik{{x}_{s}}}}}{2\left( {{e}^{2ik}}-1 \right)}\left( \begin{align}
  & \left( {{{\hat{u}}}_{-}}\left( -ik,0 \right)+{{{\hat{u}}}_{-}}\left( ik,0 \right) \right)\left( {{e}^{2ik{{x}_{s}}}}-{{e}^{2ik}} \right)+\left( {{{\hat{u}}}_{+}}\left( -ik,0 \right){{e}^{2ik}}+{{{\hat{u}}}_{+}}\left( ik,0 \right) \right)\left( {{e}^{2ik{{x}_{s}}}}-1 \right) \\
 & +\left( {{{\hat{u}}}_{-}}\left( -ik,t \right)+{{{\hat{u}}}_{-}}\left( ik,t \right) \right)\left( {{e}^{2ik}}-{{e}^{2ik{{x}_{s}}}} \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}+\left( {{{\hat{u}}}_{+}}\left( -ik,t \right){{e}^{2ik}}+{{{\hat{u}}}_{+}}\left( ik,t \right) \right)\left( 1-{{e}^{2ik{{x}_{s}}}} \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}} \\
\end{align} \right)$$
All terms involving a factor of ${{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}$ become analytic and bounded in the half planes containing their contours.  Their contribution along their contours therefore becomes zero for arguments similar to those used previously.  Hence we have effectively the following for the j-boundary data,
$${{\tilde{j}}_{+1}}\left( {{k}^{2}} \right)+ik{{\tilde{j}}_{+0}}\left( {{k}^{2}} \right)=\frac{{{e}^{ik{{x}_{s}}}}}{\left( {{e}^{2ik}}-1 \right)}\left( {{{\hat{u}}}_{-}}\left( -ik,0 \right)+{{{\hat{u}}}_{-}}\left( ik,0 \right)+{{{\hat{u}}}_{+}}\left( -ik,0 \right){{e}^{2ik}}+{{{\hat{u}}}_{+}}\left( ik,0 \right) \right)$$
This however has a factor of ${{e}^{ik{{x}_{s}}}}$ which will cancel the ${{e}^{-ik{{x}_{s}}}}$ term in ${{e}^{ikx-ik{{x}_{s}}-\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}$ within the solution essentially obliterating the boundary.  That is the solution is independent of ${{x}_{s}}$.   This is a problem! However it is easily explaind.  Observe that we may write the PDE in the following way
$\frac{\partial {{U}_{\pm }}}{\partial t}=\frac{{{\partial }^{2}}{{U}_{\pm }}}{\partial {{x}^{2}}},\,\,{{U}_{\pm }}={{u}_{\pm }}{{e}^{{{\lambda }_{\pm }}t}}$
The conditions we are using above effectively imply on the boundary that ${{U}_{+}}\left( {{x}_{s}},t \right)={{U}_{-}}\left( {{x}_{s}},t \right),\,\,{{U}_{x+}}\left( {{x}_{s}},t \right)={{U}_{x-}}\left( {{x}_{s}},t \right)$ , which explains why the boundary point is disappearing.
My assumption is that i am either missing something very obvious (regarding the physical nature of the system i.e. should we even expect this thing to have a solution, or perhaps we should be expecting a weak solution), made a mistake with the solution of the system (I have checked most of it with mathematica - so probably not), or this method can't solve this composite material type of problem (very unlikely).  My guess at this stage is that we perhaps need Robin boundary conditions on the inetermediate layer - for note that such conditions will also allow us to elminate two unknowns.  That was my next step.
Also note that book by M. Necati Ozisik "Boundary value problems of heat conduction" contains various composite material solutions, but interestingly none with two Neumann conditions on the end of the 'slab'.
Edit:
I am going to upgrade my assumption to that being the system as stated is ill-posed.  I say this because without the $\lambda $ coeff's the composite system is tractable via this method using the standard boundary conditions at the interface (and so to, i assume, via the standard classical approaches).  The issue boils down to the following four functions
$$\int\limits_{0}^{t}{\left( {{u}_{-x}}\left( {{x}_{s}},y \right)+ik{{u}_{-}}\left( {{x}_{s}},y \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}dy}={{j}_{1-}}\left( {{k}^{2}},{{\lambda }_{-}} \right)+{{j}_{0-}}\left( {{k}^{2}},{{\lambda }_{-}} \right)$$
$$\int\limits_{0}^{t}{\left( {{u}_{+x}}\left( {{x}_{s}},y \right)+ik{{u}_{+}}\left( {{x}_{s}},y \right) \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}={{j}_{1+}}\left( {{k}^{2}},{{\lambda }_{+}} \right)+{{j}_{0+}}\left( {{k}^{2}},{{\lambda }_{+}} \right)$$
What is required is a boundary condition that relates the various u+/- functions so that we can eliminate two of the transformed functions (the j-functions) to obtain a set of four unknowns for the four equations.  The various ways of doing this (as discussed previously in this post) ultimately remove the intermediate layer from the problem. The standard robin-like boundary conditions used in Ozisik for composite media won't work.  For example, consider (Ozisik)
$${{u}_{-x}}\left( {{x}_{s}},t \right)=h\left( {{u}_{-}}\left( {{x}_{s}},t \right)-{{u}_{+}}\left( {{x}_{s}},t \right) \right)$$  and
$${{u}_{-x}}\left( {{x}_{s}},t \right)={{u}_{+x}}\left( {{x}_{s}},t \right)$$
Where ‘h’ represents a contact conductance
From this we have then
$${{\hat{u}}_{-}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)t}}-{{\hat{u}}_{-}}\left( -ik,0 \right)=\int\limits_{0}^{t}{\left( -h{{u}_{+}}\left( {{x}_{s}},t \right)+\left( ik+h \right){{u}_{-}}\left( {{x}_{s}},y \right) \right){{e}^{-ik{{x}_{s}}+\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}-ik{{u}_{-}}\left( 0,y \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{-}} \right)y}}dy}$$
$${{\hat{u}}_{+}}\left( -ik,t \right){{e}^{\left( {{k}^{2}}+{{\lambda }_{+}} \right)t}}-{{\hat{u}}_{+}}\left( -ik,0 \right)=\int\limits_{0}^{t}{ik{{u}_{+}}\left( 1,y \right){{e}^{-ik+\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}-\left( h{{u}_{-}}\left( {{x}_{s}},t \right)+\left( ik-h \right){{u}_{+}}\left( {{x}_{s}},y \right) \right){{e}^{-ik{{x}_{s}}+\left( {{k}^{2}}+{{\lambda }_{+}} \right)y}}dy}$$
Note that this system is still not well-posed because we still cannot eliminate two unknowns (due to the fact that the integral transforms of the boundary value data are different transforms due to the different $\lambda $ coefficients).  If, however, the $\lambda $ coefficients were equal then it is well posed.  So if we remove the $\lambda $coefficients we can eliminate two of the unknowns and the final solution is one where the interface is present (I won’t provide the calculations unless requested, but it becomes apparent reasonably quickly using Mathematica to do the algebra).
Where to from here depends upon exactly what you are trying to model.


TITLE: Learning algebra and harmonic analysis
QUESTION [25 upvotes]: I've revised my question a bit in response to the (very helpful) advice so far--
I have an engineering background but am interested in learning abstract harmonic analysis. My interest is rather unstructured; the Fourier transform sparked my interest in mathematics, and I've been following since I first learned of it as a sort of guide through mathematics. I've decided it's time I delved into the abstract, and from my readings so far it's become clear to me that this will require a sojourn into a swathe of unfamiliar mathematics. My background in classical analysis, linear algebra, and probability is rather good (probably the rough equivalent of a 1st year american grad student), and so I'm familiar as well with point-set topology. I'm also familiar with measure theory (via probability, and some real analysis). I have essentially no abstract algebra.
I would be interested in three bits of advice:

What algebra is essential for the canon of harmonic analysis? Perhaps this is a vacuous question (i.e., that an answer is as simple "just google" or "open a book")? It strikes me that techniques or background might work their way in unnamed. I'm hoping to prevent the sinking feeling that I don't know something I should.
Specific examples (theorems, objects, counterexamples) which illustrate important applications of harmonic analysis, especially in relation to algebra. I'm not looking for explanations, just statements of what's thought to be important. I'm hoping that this information will help me sift through the different branches of harmonic analysis, especially where I am most unfamiliar with the language.
Canonical/important texts which I might find useful. I'm especially interested in those good for self study (well-edited with many examples and end-of-chapter exercises, and ideally a conversational tone with lengthy remarks and some historical content).

REPLY [18 votes]: Let me suggest that you look at the book "An introduction to topology and modern analysis", by Simmons.  It covers concepts of point set topology that you presumably already know, but also gives a fairly concise, but quite readable, introduction to abstract algebra.  It then brings these together in its final sections into a discussion of Banach algebras and related topics, culminating in a proof of the Gelfand--Naimark theorem.  [See below for a brief remark about this theorem, and the other theorems mentioned in the subsequent paragraphs.]
Since you say that you are interested in abstract harmonic analysis, the Gelfand--Naimark theorem is a good place to start.  (For example, it is not so far to go from there to the abstact form of Wiener's Tauberian theorem.)
Note: I am interpreting abstract harmonic analysis to mean something like harmonic analysis on locally compact abelian groups (and related topics).
Simmons also has exercises, I think.
When I was studying this stuff, the next place I went to after Simmons was Naimark's tome Normed rings.  (There are various editions, and some of the later ones might be called Normed algebras instead, if I'm not misremembering.  They are translated from Russian, so the slighly unusual, and changing, name may be an artefact of this; I'm not sure.  In any case, they are basically about the theory of Banach algebras and its applications to abstract harmonic analysis.)
This is a place where one can read about various group rings of topological groups, Haar measure, the general form of Wiener's Tauberian theorem, and other concepts of abstract harmonic analysis.  It is essentially too long to read from
start to finish, but in my experience one can dip into it in bits and pieces, and having a firm understanding of the material from Simmons helps a lot.
Naimark's book is a monograph, not a textbook as such, and although it has many historical comments and illustrative examples (although the examples are often at a theoretically fairly high level), I don't remember it as having exercises.  But in any case, I am not suggesting it as a first point of call, but as somewhere to go after you have some basics under your belt.
There is also a book by Loomis, An introduction to abstract harmonic analysis,
which also treats Haar measure, various group rings, and so on.  If I remember correctly it is  less condensed than Naimark and also less comprehensive.  My memory is that I preferred Naimark, but probably for idiosyncratic reasons.  I don't remember whether Loomis's book has exercises.
All the books I'm mentioning are probably out of print, so I'm also assuming that you have access to a university library or something similar.  (Any decent such library should have them.)
Finally, some fundamental results that I would recommend aiming for, which combine algebra
and analysis nicely:

Gelfand's generalization of Wiener's theorem (that if $f$ is a nowhere zero continuous periodic function whose Fourier series is absolutely convergent, than
the Fourier series of $1/f$ is also absolutely convergent).
the Gelfand--Naimark theorem (identifying certain commutative Banach algebras with extra
structure as being algebras of continuous functions on compact topological space; it is a beautiful generalization of the classical spectral theorems for matrices).
The generalization of Wiener's Tauberian theorem to arbitrary commutative locally compact groups.  (Wiener's original theorem says that if $f$ is an $L^1$-function on the real line whose Fourier transform is nowhere zero, then
the translates of $f$ span a dense subspace of $L^1$.)
More abstract, but basic to the previous example and lots of other things, is
the existence of Haar measure for any locally compact group.

As already mentioned,
the first two results are in Simmons, and are easier, but already involve a very nice interplay between analysis and algebra.  (The general framework is that of Banach algebras, which combine the analysis of Banach spaces with the
algebra of rings, ideals, and so on.)
The second two results are in Naimark, and Haar measure is also in Loomis (and many other places) (and the general form of Wiener may be in Loomis too; I forget now).