\documentclass[12pt, a4paper, twoside]{article}

\usepackage{amsmath, amsthm, amsfonts, array}
\usepackage {amssymb, amscd, latexsym, times, epic, eepic}

\title{The complex Riccati equation}
\author{Matias Dahl}
\date{\today}

\parindent=0cm
\parskip=.2cm

\theoremstyle{plain}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}

\newcommand {\Set}[1] {\mathbb{#1}}
\newcommand {\slaz}[0] {\backslash\{0\}}
\newcommand {\od}[2] {\frac{d#1}{d{#2}}}
\newcommand {\IM}[0] {\operatorname{Im}}
\newcommand {\sC}[0] {\Set{C}}
\newcommand {\sR}[0] {\Set{R}}


\begin{document}

\maketitle

In the below, all matrices are assumed to be complex 
  $n\times n$ matrices unless otherwise mentioned. 
By a positive definite matrix we mean a real symmetric matrix $A$ 
  such that $\eta^T \cdot A \cdot \eta > 0 $ for all $\eta\in \sR^n\slaz$,
or equivalently, if
  $\eta^T \cdot A \cdot \eta > 0 $ for all $\eta\in \sC^n\slaz$.
If $A$ is a matrix we denote the transpose, 
  conjugate transpose, and imaginary part of $A$ 
  by $A^T$, $A^\ast$, and $\IM A$, respectively. 
By $\langle \ ,\ \rangle$ we denote the usual inner product 
   in $\sC^n$; for $\eta=(\eta_1, \ldots, \eta_n)$, 
    $\zeta=(\zeta_1, \ldots, \zeta_n) \in \sC^n$, 
   $$
    \langle\eta,\zeta \rangle = \sum_{k=1}^n \eta_k \overline{\zeta_k}.
   $$

Suppose $I\ni 0$ is an open interval, and suppose $B,C,D$ are real matrices 
  depending continuously on $t\in I$, suppose $C,D$ are symmetric, and 
  all matrices are bounded on $I$.
Under these assumptions, we shall study the \emph{Riccati equation}
  \begin{eqnarray}
  \label{ricrep}
    \od{H}{t} + BH + HB^T + HCH + D = 0.
  \end{eqnarray}

Following \cite{KKL}, we shall prove the following result:

\begin{proposition}
\label{appresultII}
Suppose $H_0$ is a symmetric $n\times n$ matrix such 
that $\IM H_0$ is positive definite. 
Then equation \ref{ricrep} has a unique solution $H$ on $I$ such that 
\begin{enumerate}
\item $H(0)=H_0$, 
\item $H$ is symmetric and $\IM H$ is positive definite for all $t$.
\end{enumerate}
\end{proposition}

To prove this we first study the linear differential equation,
\begin{eqnarray}
\label{diffeq}
\od{}{t}\begin{pmatrix}Y \\ Z\end{pmatrix} &=& \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} \begin{pmatrix} D & B \\ B^T & C \end{pmatrix}\begin{pmatrix}Y \\ Z\end{pmatrix}, \\
\label{diffeqin}
\begin{pmatrix}Y \\ Z\end{pmatrix}\Big|_{t=0} &=& \begin{pmatrix}Y_0 \\ Z_0\end{pmatrix}.
\end{eqnarray}
Here $Y,Z$ are unknown matrices depending on $t\in I$, and 
  the initial value $(Y_0,Z_0)$ is known.
In \cite{taylor} it is proven that equation \ref{diffeq} has a 
  global solution on $I$ and this solution is uniquely determined 
  by the initial condition $(Y_0,Z_0)$. 

\begin{lemma}
\label{Hinvariant}
Suppose $Y_0,Z_0$ are matrices such that $Y_0$ is invertible,  
  $Z_0 Y_0^{-1}$ is symmetric, 
  and $\IM (Z_0 Y_0^{-1})$ is positive definite. 
Further, suppose $(Y,Z)$ is the solution to equations \ref{diffeq},
\ref{diffeqin} 
   with initial value $(Y_0, Z_0)$.  
Then matrices 
\begin{eqnarray*}
Z^T Y - Y^T Z,\quad\quad Z^\ast Y - Y^\ast Z
\end{eqnarray*}
are independent of $t \in I$, and 
   $Y$ is invertible for all $t$. 
\end{lemma}

\newcommand{\dtwo}[2]{\begin{pmatrix} #1 & #2 \end{pmatrix} }
\newcommand{\utwo}[2]{\begin{pmatrix} #1 \\ #2 \end{pmatrix} }

\begin{proof}
 Let $\tau\in \{T, \ast\}$. Since $(Y,Z)$ satisfies 
   equation \ref{diffeq}, and $B,C,D$ are real matrices, we have
\begin{eqnarray*}
\od{}{t}\left( \utwo{Y}{Z}^\tau \begin{pmatrix}0 & -I \\ I & 0 \end{pmatrix} \utwo{Y}{Z}\right) = 0,
\end{eqnarray*}
and the first two claims follows. 
Let us next show that $Y$ is invertible for all $t\in I$.  
   For a contradiction, suppose $Y(s)\eta=0$ for some 
   $s\in I$, $\eta\in \sC^n$. Then
\begin{eqnarray*}
\langle (Z_0^\ast Y_0 - Y_0^\ast Z_0)\eta, \eta\rangle 
&=&\langle (Z^\ast Y - Y^\ast Z)(s) \eta, \eta\rangle \\
&=& 0.
\end{eqnarray*}
For a complex number $\alpha \in \sC$, we have 
   $\IM \alpha = \frac{i}{2}(\overline{\alpha}-\alpha)$, so
for any $\zeta \in \sC^n$, 
\begin{eqnarray*}
\langle \IM (Z_0 Y_0^{-1}) \zeta, \zeta \rangle  %&=& \frac{i}{2} \langle (H_0^\ast-H_0) \zeta, \zeta \rangle \\
   &=& \frac{i}{2} \langle (Z_0^\ast Y_0 - Y_0^\ast Z_0 ) Y_0^{-1} \zeta, Y_0^{-1} \zeta \rangle.
\end{eqnarray*}
Putting $\zeta = Y_0 \eta$ gives 
   a contradiction with the assumption on $\IM(Z_0 Y_0^{-1})$.
\end{proof}

The next lemma shows how equations \ref{ricrep} and \ref{diffeq} 
   are related. 

\begin{lemma} 
\label{equequiv}
$\, $
\begin{enumerate}
\item Suppose $Y_0,Z_0, Y,Z$ are as in Lemma \ref{Hinvariant}.
Then 
$$
   H = Z Y^{-1}
$$
is a solution to equation \ref{ricrep} for all $t\in I$
with initial value $H(0)=Z_0 Y^{-1}_0$. 
\item Let $H$ be a solution to equation \ref{ricrep} with 
  initial value $H_0$. Then there exist
  matrices $Y_0,Z_0$ such that 
  $Y_0$ is invertible, 
$$
   H_0 = Z_0 Y^{-1}_0,
$$
and
$$
   H = Z Y^{-1}, 
$$
when $(Y,Z)$ is the solution to equations \ref{diffeq}, \ref{diffeqin},
  with initial value $(Y_0,Z_0)$.
\end{enumerate}
\end{lemma}

\begin{proof} 
For any matrix invertible and differentiable for all $t\in I$, we have 
$$
   \od{A^{-1}}{t} = -A^{-1} \od{A}{t}A^{-1}.
$$
Thus, if $(Y,Z)$ is a solution to equation \ref{diffeq}, then
$$
  \od{}{t}(ZY^{-1}) = -BH-HB^T - HCH -D.
$$
This formula implies the first claim. 
In the second claim, $Y_0=I$, $Z_0 = H_0$ is a suitable choice of 
  $Y_0, Z_0$. 
The representation $H=ZY^{-1}$ follows since 
$(H-ZY^{-1})(0)=0$, and 
$
  \od{}{t}(H-ZY^{-1})=0.
$
\end{proof}

\begin{proof}[Proof of Proposition \ref{appresultII}.]
Since $H_0 = Z_0 Y_0^{-1}$ for the choice $Y_0=I$, $Z_0 = H_0$, 
   Lemma \ref{equequiv}.1 implies that 
   equation \ref{ricrep} has a global solution. 
For uniqueness, suppose $H,\tilde{H}$ are two solutions to 
   equation \ref{ricrep} satisfying
$$
  H(0)=\tilde{H}(0)=H_0.
$$
Then Lemma \ref{equequiv}.2 implies that there are matrices
 $Y_0, \tilde{Y}_0, Z_0, \tilde{Z}_0$ such that 
\begin{eqnarray}
\label{eqqww}
  H_0 = Z_0 Y^{-1}_0 = \tilde{Z}_0 \tilde{Y}^{-1}_0, 
\end{eqnarray}
and
$$
  H = ZY^{-1}, \quad \tilde{H} = \tilde{Z} \tilde{Y}^{-1},
$$
where $(Y,Z)$ and $(\tilde{Y}, \tilde{Z})$ are the solutions
to equation \ref{diffeq} with initial values $(Y_0, Z_0)$, 
$(\tilde{Y}_0, \tilde{Z}_0)$, respectively. 
We know that solutions to equation \ref{diffeq} are uniquely determined
  by the initial value. Thus 
\begin{eqnarray}
\label{eqqwww}
  \tilde{Y} = Y (Y_0^{-1} \tilde{Y}_0), \quad   \tilde{Z} = Z (Y_0^{-1} \tilde{Y}_0).
\end{eqnarray}
Indeed, as $(Y_0^{-1} \tilde{Y}_0)$ is a constant matrix, 
   the right hand sides in equation \ref{eqqwww} satisfy equation \ref{diffeq}
   and the same initial condition as $(\tilde{Y},\tilde{Z})$. 
By equation \ref{eqqww},
$$ 
  \tilde{H} = \tilde{Z} \tilde{Y}^{-1} = Z Y^{-1} = H,
$$
and the solution to equation \ref{ricrep} is unique.
Suppose $H$ is a solution to equation \ref{ricrep}. Then $H^T$ is also
a solution to equation \ref{ricrep}, and since $H$ and $H^T$ share the 
same initial value, $H$ is symmetric. 
It remains to prove that $\IM H$ is positive definite. 
Since $\IM H_0$ is symmetric, it follows that
    $\IM H_0 = \frac{i}{2} (H_0^\ast-H_0)$, 
so
$$  
\IM H_0 = \frac{i}{2}(Y_0^{-1})^\ast ( Z_0^\ast Y_0- Y_0^\ast Z_0) Y_0^{-1}.
$$             
and since $\IM H$ is symmetric, 
\begin{eqnarray*}
  \IM H &=& \frac{i}{2}(Y^{-1})^\ast ( Z^\ast Y-Y^\ast Z) Y^{-1} \\
        &=& (Y_0 Y^{-1})^\ast \IM (H_0) (Y_0 Y^{-1})
\end{eqnarray*}
by Lemma \ref{Hinvariant}.
\end{proof}

\begin{thebibliography}{9}

\bibitem{KKL} A.~Kachalov, Y.~Kurylev, M.~Lassas, 
   \emph{Inverse Boundary Spectral Problems}, 
   Chapman \& Hall/CRC, 2001.

\bibitem{taylor}
    M.~E.~Taylor,
    \emph{Partial Differential Equations: Basic Theory},
    Springer, 1996.
\end{thebibliography}





\end{document}


