\section{Introduction}

Majumdar\cite{maj47} and Papapetrou\cite{papa47} studied charged
dusts in the light of general relativity, and later also by many
others (see Ray et al.\cite{remlz03} for a detailed reference).
Charged fluid spheres have been studied by Bekenstein\cite{bek71},
Bonnor\cite{bon80}, Zhang et al.\cite{zhang82}, etc. Zhang et
al.\cite{zhang82} indirectly verified that the structure of a
star, for a degenerate relativistic fermi gas, is significantly
affected by the electric charge  just when the charge density is
close to the mass density.

We took trapped protons to be the charge carriers in the star. The
effect of charge does not depend on its sign by our formulation.
The energy density which appears from the electrostatic field
will {\it add up} to the total energy density of the system,
which in turn will  help in the {\it gaining} of the  total mass
of  the system. The modified Tolman-Oppenheimer-Volkoff (TOV)
equation now has extra terms due to the presence of  the
Maxwell-Einstein stress tensor. We solve the modified TOV
equation for  polytropic EOS assuming that the charge density
goes with the matter density.

The detailed relations are shown in Ray et al.\cite{remlz03}. We
used the modified TOV and see the effect of charge on a model
independent polytropic EOS. We assumed the  charge is proportional
to the mass density ($\epsilon$) like $\rho_{ch}=f \times
\epsilon $ where $\epsilon=\rho c^2$ is in [MeV/fm$^3$].

The polytropic EOS is given by $P=\kappa \rho^{1+1/n} $ where $n$
is the  polytropic index and for our present choice of EOS, we
took $n=1.5$ and $\kappa=0.05[fm]^{8/3}$.

\begin{figure}[ht]
\centerline{\epsfxsize=5cm\epsfbox{m-rhoc.eps}} \caption{Central
density against mass for different values of the factor
$f$.\label{fig:poly-e-m}}
\end{figure}

In  Fig.(\ref{fig:poly-e-m}), we plot the mass as function of the
central density, for  different values  of the  charge fraction
$f$. For the charge fraction $f=0.0001$, we do not see any
departure on the stellar structure from that of the chargeless
case. This value of $f$ is $critical$ because any increase in the
value beyond this, shows enormous effect on the structure. The
increase of the maximum mass of the star is very much non-linear,
as can be seen from the Fig. (\ref{fig:poly-e-m}).

\begin{figure}[htbp]
\centerline{\psfig{figure=q-m-poly-n15.eps,width=5cm}}
\caption{\label{fig:poly-q-m}The variation of the charge with
mass (right) for different $f$.}
\end{figure}

The Q$\times$ M diagram of Fig.(\ref{fig:poly-q-m}) shows the
mass of the stars against their surface charge. We have made the
charge density proportional to  the energy  density and so  it
was expected  that the charge, which is  a volume integral of the
charge  density, will go in the same  way as the  mass, which is
also a volume integral  over the mass density.  The slope of the
curves comes from the different charge fractions. If we consider
that the maximum allowed charge estimated by the condition ($U
\simeq \sqrt{8\pi P} < \sqrt{8\pi\epsilon}$) for $\frac{dP}{dr}$
to be negative (Eq.~12 of Ray et al.\cite{remlz03})),  we see that
the curve for the maximum charge  in Fig.(\ref{fig:poly-q-m}) has
a slope of 1:1 (in a charge scale of 10$^{20}$
Coulomb\cite{remlz03}). This scale can easily be understood if we
write the charge as
$Q=\sqrt{G}M_\odot\frac{M}{M_\odot}\simeq10^{20}\frac{M}{M_\odot}{\rm
Coulomb}.$ This charge Q  is the charge at  the surface of the
star where the  pressure and  also $\frac{dP}{dr}$  are zero.

The total  mass of the star increases with increasing charge
because the  electric energy density $adds~on$ to  the mass energy
density. This change in  the mass is low for smaller charge
fraction  and going  up to  12 times  the value  of chargeless
case  for maximum allowed charge  fraction $f=0.0011$. The most
effective term   in  Eq.~12 of Ray et al.\cite{remlz03} is  the
factor ($\rm M_{tot}+4\pi r^3P^*$). $P^*= P-\frac{U^2}{8\pi}$ is
the effective pressure of the system because the effect of charge
decreases the outward fluid pressure, negative in sign to the
inward gravitational pressure. With the increase of charge, P$^*$
decreases, and hence the gravitational negative part of Eq.~12 of
Ray et al.\cite{remlz03} decreases. So, with the softening of the
pressure  gradient, the system allows more radius for the star
until  it reaches  the surface where the pressure (and
$\frac{dP}{dr}$) goes to zero. Since $\frac{U^2}{8\pi}$ cannot
be  too much larger than the pressure in order to maintain
$\frac{dP}{dr}$  negative, so we have a limit  on  the charge,
which comes from the relativistic effects of the gravitational
force and not just only from the repulsive Coulombian part.

In our study, we have shown that a high density system like a
neutron star can hold huge charge of the order of 10$^{20}$
Coulomb considering the global balance of forces. With the
increase of charge, the maximum mass of the star recedes back to
a lower density regime. The stellar mass also increases rapidly
in the critical limit of the maximum charge content, the systems
can hold. The radius also increases accordingly, however keeping
the M/R ratio increasing with charge. The increase in mass is
primarily brought in by the softening of the pressure gradient due
to the presence of a Coulombian term coupled with the
Gravitational matter part. Another intrinsic increase in the mass
term comes through the addition of the electric energy density to
the mass density of the system. The stability of these charged
stars are however ruled out from the consideration of forces
acting on individual charged particles. They face enormous radial
repulsive force and leave the star in a very short time. This
creates an imbalance of forces and the gravitational force
overwhelms the repulsive Coulomb and fluid pressure forces and
the star collapses to a charged black hole. Finally, these
charged stars are supposed to be very short lived, and are the
intermediate state between a supernova collapse and charged black
holes\cite{remlz03}.


\section{Introduction}
\label{sec:intro}

An airborne hyperspectral imaging sensor is capable of simultaneously acquiring the same spatial scene in a contiguous and multiple narrow (0.01 - 0.02 $\mu$m) spectral wavelength (color) bands \cite{Shaw02, manolakis2003hyperspectral, manolakis_lockwood_cooley_2016, 974715, 7564440, ZHANG20153102, 6555921, PLAZA2009S110, 6509473, bookJocekynToaPPEAR, DallaMura2011, prasad2011optical, chanussotDec2009}. When all the spectral bands are stacked together, the result is a hyperspectral image (HSI) whose cross-section is a function of the spatial coordinates and its depth is a function of wavelength. Hence, an HSI is a 3-D data cube having two spatial dimensions and one spectral dimension. Thanks to the narrow acquisition, the HSI could have hundreds to thousands of contiguous spectral bands. Having this very high level of spectral detail gives better capability to see the unseen.
\\
Each band of the HSI corresponds to an image of the surface covered by the field of view of the hyperspectral sensor, whereas each ``pixel'' in the HSI is a $p$-dimensional vector, $\mathbf{x}\in\mathbb{R}^p$ ($p$ stands for the total number of spectral bands), consisting of a spectrum characterizing the materials within the pixel. The ``spectral signature'' of $ \mathbf{x}$ (also known as ``reflectance spectrum''), shows the fraction of incident energy, typically sunlight, that is reflected by a material from the surface of interest as a function of the wavelength of the energy \cite{JoanaThesis}.
\\
The HSI usually contains both pure and mixed pixels \cite{manolakis2003hyperspectral, Manolakis09, 6048900, 6504505, 5658102, 6351978}. A pure pixel contains only one single material, whereas a mixed pixel contains multiple materials, with its spectral signature representing the aggregate of all the materials in the corresponding spatial location. The latter situation often arises because hyperspectral images are collected hundreds to thousands of meters away from an object so that the object becomes smaller than the size of a pixel. Other scenarios might involve, for example, a military target hidden under foliage or covered with camouflage material.
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With the rich information afforded by the high spectral dimensionality, hyperspectral imagery has found many applications in various fields, such as astronomy, agriculture \cite{Patel2001, Datt2003}, mineralogy \cite{Lehmann2001}, military \cite{manolakis2002detection, Stein02, 4939406}, and in particular, target detection \cite{Shaw02, manolakis2003hyperspectral, Manolakis14, Manolakis09, manolakis2002detection, 7739987, 7165577, 8069001, Frontera14, 6378408}. Usually, the detection is built using a binary hypothesis test that chooses between the following competing null and alternative hypothesis: target absent ($H_0$), that is, the test pixel $\mathbf{x}$ consists only of background, and target present ($H_1$), where $\mathbf{x}$ may be either fully or partially occupied by the target material.
\\
It is well known that the signal model for hyperspectral test pixels is fundamentally different from the additive model used in radar and communications applications \cite{manolakis_lockwood_cooley_2016, Manolakis09}. We can regard each test pixel $\mathbf{x}$ as being made up of $\mathbf{x} = \alpha \, \mathbf{t}$ + $(1-\alpha) \, \mathbf{b}$, where $0 \leq \alpha \leq 1$ designates the target fill-fraction, $\mathbf{t}$ is the spectrum of the target, and $\mathbf{b}$ is the spectrum of the background. This model is known as replacement signal model, and hence, when a target is present in a given HSI, it replaces (that is, removes) an equal part of the background \cite{manolakis_lockwood_cooley_2016}. For notational convenience, sensor noise has been incorporated into the target and background spectra; that is, the vectors $\mathbf{t}$ and $\mathbf{b}$ include noise \cite{manolakis_lockwood_cooley_2016}.
\\
In particular, when $\alpha=0$, the pixel $\mathbf{x}$ is fully occupied by the background material (that is, the target is not present). When $\alpha = 1$, the pixel $\mathbf{x}$ is fully occupied by the target material and is usually referred to as the full or resolved target pixel. Whereas when $0<\alpha<1$, the pixel $\mathbf{x}$ is partially occupied by the target material and is usually referred to as the subpixel or unresolved target \cite{Manolakis09}.
\\
\\
A prior target information can often be provided to the user. In real world hyperspectral imagery, this prior information may not be only related to its spatial properties (e.g.  size, shape, texture) and which is usually not at our disposal, but to its spectral signature. The latter usually hinges on the nature of the given HSI where the spectra of the targets of interests have been already measured by some laboratories or with some hand-held spectrometers.
\\
Different Gaussian-based target detectors (e.g. Matched Filter \cite{Manolakis00, Nasrabadi08}, Normalized Matched Filter \cite{kraut1999cfar}, and Kelly detector \cite{Kelly86}) have been developed. In these classical detectors, the target of interest to detect is known, that is, its spectral signature is fully provided to the user. However, these detectors present several limitations in real-world hyperspectral imagery. The task of understanding and solving these limitations presents significant challenges for hyperspectral target detection.
\begin{itemize}
\item{\bf Challenge one:} One of the major drawbacks of the aforementioned classical target detectors is that they depend on the unknown covariance matrix (of the background surrounding the test pixel) whose entries have to be carefully estimated, especially in large dimensions \cite{LEDOIT2004365, LedoitHoney, AhmadCamsap2017}, and to ensure success under different environment \cite{5606730, 6884641, 6894189}. However, estimating large covariance matrices has been a longstanding important problem in many applications and has attracted increased attention over several decades.
When the spectral dimension is considered large compared to the sample size (which is the usual case), the traditional covariance estimators are estimated with a lot of errors unless some covariance regularization methods are considered \cite{LEDOIT2004365, LedoitHoney, AhmadCamsap2017}. It implies that the largest or smallest estimated coefficients in the matrix tend to take on extreme values not because this is ``the truth'', but because they contain an extreme amount of error \cite{LEDOIT2004365, LedoitHoney}. This is one of the main reasons why the classical target detectors usually behave poorly in detecting the targets of interests in a given HSI.
\\
In addition, there is always an explicit assumption (specifically, Gaussian) on the statistical distribution characteristics of the observed data. e statistical distribution characteristics of the observed data. For instance, most materials are treated as Lambertian because their bidirectional reflectance distribution function characterizations are usually not available, but the actual reflection is likely to have both a diffuse component and a specular component. This latter component would result in gross corruption of the data. In addition, spectra from multiple materials are usually assumed to interact according to a linear mixing model; nonlinear mixing effects are not represented and will contribute to another source of noise
\\
\item{\bf Challenge two:} The classical target detectors that depend on the target to detect $\mathbf{t}$, use only a single reference spectrum for the target of interest. This may be inadequate since in real world hyperspectral imagery, various effects that produce variability to the material spectra (e.g. atmospheric conditions, sensor noise, material composition, and scene geometry) are inevitable \cite{803418, 1000320}.
For instance, target signatures are typically measured in laboratories or in the field with handheld spectrometers that are at most a few inches from the target surface. Hyperspectral images, however, are collected at huge distances away from the target and have significant atmospheric effects present.
\end{itemize}
Recent years have witnessed a growing interest on the notion of sparsity as a way to model signals. The basic assumption of this model is that natural signals can be represented as a ``sparse'' linear combination of atom signals taken from a dictionary. In this regard, two main issues need to be addressed: 1) how to represent a signal in the sparsest way, for a given dictionary and 2) how to construct an accurate dictionary in order to successfully representing the signal.
\\
 Recently, a signal classification technique via sparse representation was developed for the application of face recognition \cite{4483511}. It is observed that aligned faces of the same object with varying lighting conditions approximately lie in a low-dimensional subspace \cite{1177153}. Hence, a test face image can be sparsely represented by atom signals from all classes. This representation approach has also been exploited in several other signal classification problems such as iris recognition \cite{5339067}, tumor classification \cite{1177153asf}, and hyperspectral imagery unmixing \cite{doi818245, 6555921, 6200362}.
\\
In this context, Chen {\it et al.} \cite{Chen11b} have been inspired by the work in \cite{4483511}, and developed an approach for sparse representation of hyperspectral test pixels. In particular, each test pixel $\mathbf{x} \in\mathbb{R}^p$ in a given HSI, be it target or background, is assumed to lie in a low dimensional subspace of the $p$-dimensional spectral-measurement space. Hence, it can be represented by a very few atom signals taken from dictionaries, and the recovered sparse representation can be used directly for detection. For example, if a test pixel $\mathbf{x}$ contains the target (that is, $\mathbf{x} = \alpha \, \mathbf{t} + (1-\alpha)\, \mathbf{b}$, with $0<\alpha \leq 1$), thus it can be sparsely represented by atom signals taken from the target dictionary (denoted as $\mathbf{A}_t)$; whereas, if $\mathbf{x}$ is only a background pixel (that is, it does not contain the target, e.g., $\alpha=0$), thus, it can be sparsely represented by atom signals taken from the  background dictionary (denoted as $\mathbf{A}_b$).
Very recently, Zhang {\it et al.} \cite{Zhang15} have extended the work done by Chen {\it et al.} in \cite{Chen11b} by combining the idea of binary hypothesis and sparse representation together, obtaining a more complete and realistic sparsity model than in \cite{Chen11b}. More precisely, Zhang {\it et al.} \cite{Zhang15} have assumed that if the test pixel $\mathbf{x}$ belongs to hypothesis $H_0$ (that is, the target is absent), it will be modeled by the $\mathbf{A}_b$ only; otherwise, it will be modeled by the union of $\mathbf{A}_b$ and $\mathbf{A}_t$. This in fact yields a competition between the two hypotheses corresponding to the different pixel class label.
\\
\\
These sparse representation methods \cite{Chen11b, Zhang15} are being independent on the unknown covariance matrix, behave well in large dimensions, distributional free and invariant to atmospheric effects. More precisely, they can alleviate the spectral variability caused by atmospheric effects, and can also better deal with a greater range of noise phenomena. The main drawback of these approaches is that they usually lack a sufficiently universal dictionary, especially for the background $\mathbf{A}_b$; some form of in-scene adaptation would be desirable. The background dictionary $\mathbf{A}_b$ is usually constructed using an adaptive scheme (that is, a local method) which is  based on a dual concentric window centered on the test pixel, with an inner window region (IWR) centered within an outer window region (OWR), and only the pixels in the OWR will constitute the samples for $\mathbf{A}_b$. Clearly, the dimension of IWR is very important and has a strong impact on the target detection performance since it aims to enclose the targets of interests to be detected. It should be set larger than or equal to the size of all the desired targets of interests in the corresponding HSI, so as to exclude the target pixels from erroneously appearing in $\mathbf{A}_b$. However, information about the target size in the image is usually not at our disposal. It is also very unwieldy to set this size parameter when the target could be of irregular shape (e.g. searching for lost plane parts of a missing aircraft). Another tricky situation is when there are multiple targets in close proximity in the image (e.g. military vehicles in long convoy formation).
\\
\\
Hence, an important third challenge appears:
\begin{itemize}
\item {\bf Challenge three:} The construction of $\mathbf{A}_b$ for the sparse representation methods is a very challenging problem since a contamination of it by the target pixels can potentially affect the target detection performance.
\end{itemize}
$\\$
In this chapter, we handle all the aforementioned challenges by making very little specific assumptions about the background or target \cite{8462257, 8677268}. Based on a modification of the recently developed Robust Principal Component Analysis (RPCA) \cite{Candes11}, our method decomposes an input HSI into a background HSI (denoted by $\mathbf{L}$) and a sparse target HSI (denoted by $\mathbf{E}$) that contains the targets of interests.
\\
While we do not need to make assumptions about the size, shape, or number of the targets, our method is subject to certain generic constraints that make less specific assumption on the background or the target. These constraints are similar to those used in RPCA \cite{Candes11, NIPS2009_3704}, including:
\begin{enumerate}
\item the background is not too heavily cluttered with many different materials with multiple spectra, so that the background signals should span a low-dimensional subspace, a property that can be expressed as the low rank condition of a suitably formulated matrix \cite{ChenYu, Zhang15b, 7322257, 8260545, 8126244, Ahmad2017a, 8126244, 8260545, 7312998};
\item the total image area of all the target(s) should be small relative to the whole image (i.e. spatially sparse), e.g., several hundred pixels in a million-pixel image, though there is no restriction on a target shape or the proximity between the targets.
\end{enumerate}
Our method also assumes that the target spectra are available to the user and that the atmospheric influence can be accounted for by the target dictionary $\mathbf{A}_t$. This pre-learned target dictionary $\mathbf{A}_t$ is used to cast the general RPCA into a more specific form, specifically, we further factorize the sparse component $\mathbf{E}$ from RPCA into the product of $\mathbf{A}_t$ and a sparse activation matrix $\mathbf{C}$ \cite{8462257}. This modification is essential to disambiguate the true targets from the background

 \begin{figure}[!tbp]
 \centering
\minipage{0.84\textwidth}
  \includegraphics[width=\linewidth]{target_scheme.png}
\endminipage
\caption{Sparse target HSI: our novel target detector.}\label{fig:image1}
\end{figure}
$\\$
After decomposing a given HSI into the sum of low-rank HSI and a sparse HSI containing only the targets with the background is suppressed, the detector is simply defined by the sparse HSI. That is, the targets are detected at the non-zero entries of the sparse HSI. Hence, a novel target detector is developed and which is simply a sparse HSI generated automatically from the original HSI, but containing only the targets with the background is suppressed (see Fig. \ref{fig:image1}).
\\
The main advantages of our proposed detector are the following: 1) independent on the unknown covariance matrix; 2) behaves well in large dimensions; 3) distributional free; 4) invariant to atmospheric effects via the use of a target dictionary $\mathbf{A}_t$ and 5) does not require a background dictionary to be constructed.
\\
\\
This chapter is structured along the following lines. First comes an overview of some related works in section \ref{sec:RelatedWorks}. In Section \ref{sec:Maincontribution}, the proposed decomposition model as well as our novel target detector are briefly outlined. Section \ref{sec:experimentss} presents real experiments to gauge the effectiveness of the proposed detector for hyperspectral target detection. This chapter ends with a summary of the work and some directions for future work.
\\
\\
{\em Summary of Main Notations:} Throughout this chapter, we depict vectors in lowercase boldface letters and matrices in uppercase boldface letters. The notation $(.)^T$ and $\mathrm{Tr}(.)$ stand for the transpose and trace of a matrix, respectively. In addition, $\mathrm{rank}(.)$ is for the rank of a matrix. A variety of norms on matrices will be used. For instance, $\mathbf{M}$ is a matrix, and $[\mathbf{M}]_{:,j}$ is the $j$th column. The matrix $l_{2,0}$, $l_{2,1}$ norms are defined by $\left\Vert\mathbf{M}\right\Vert_{2,0} = \# \left\{ j \, : \, \left\Vert\left[\mathbf{M}\right]_{:,j}\right\Vert_2 \, \not= \, 0\right\}$, and $\left\Vert\mathbf{M}\right\Vert_{2,1} = \sum_{j} \left\Vert\left[\mathbf{M}\right]_{:,j}\right\Vert_2$, respectively. The Frobenius norm and the nuclear norm (the sum of singular values of a matrix) are denoted by $\left\Vert\mathbf{M}\right\Vert_F$ and $\left\Vert\mathbf{M}\right\Vert_* = \Tr\left(\mathbf{M}^T \,\mathbf{M}\right)^{(1/2)}$, respectively.

\section{Related works}
\label{sec:RelatedWorks}
Whatever the real application may be, somehow the general RPCA model needs to be subject to further assumptions for successfully distinguishing the true targets from the background.
Besides the generic RPCA and our proposed modification discussed in section \ref{sec:intro}, there have been other modifications of RPCA.
For example, the generalized model of RPCA, named the Low-Rank Representation (LRR) \cite{6180173}, allows the use of a subspace basis as a dictionary or just uses self-representation to obtain the LRR. The major drawback in LRR is that the incorporated dictionary has to be constructed from the background and to be pure from the target samples. This challenge is similar to the aforementioned background dictionary $\mathbf{A}_b$  construction problem. If we use the self-representation form of LRR, the presence of a target in the input image may only bring about a small increase in rank and thus be retained in the background \cite{8677268}.
\\
\\
In the earliest models using a low rank matrix to represent background \cite{Candes11, NIPS2009_3704, SPCPCandes}, no prior knowledge on the target was considered. In some applications such as Speech enhancement and hyperspectral imagery, we may expect some prior information about the target of interest, which can be provided to the user. Thus, incorporating this information about the target into the separation scheme in the general RPCA model should allow us to potentially improve the target extraction performance. For example, Chen and Ellis \cite{6701883} and Sun and Qin \cite{7740039} proposed a Speech enhancement system by exploiting the knowledge about the likely form of the targeted speech. This was accomplished by factorizing the sparse component from RPCA into the product of a dictionary of target speech templates and a sparse activation matrix. The proposed methods in \cite{6701883} and \cite{7740039} typically differ on how the fixed target dictionary of speech spectral templates is constructed.
Our proposed model in section \ref{sec:Maincontribution} is very related to \cite{6701883} and \cite{7740039}. In real-world hyperspectral imagery, the prior target information may not be only related to its spatial properties (e.g.  size, shape, and texture), which is usually not at our disposal, but to its spectral signature. The latter usually hinges on the nature of the given HSI where the spectra of the targets of interests present have been already measured by some laboratories or with some handheld spectrometers.
\\
In addition, by using physical models and the MORTRAN atmospheric modeling program \cite{Berk89}, a number of samples for a specific target can be generated under various atmospheric conditions.

\section{Main Contribution}
\label{sec:Maincontribution}

Suppose an HSI of size $h \times w \times p$, where $h$ and $w$ are the height and width of the image scene, respectively, and $p$ is the number of spectral bands. Our proposed modification of RPCA is mainly based on the following steps:
\begin{enumerate}
\item Let us consider that the given HSI contains $q$ pixels $\{ \mathbf{x}_i\}_{i\in[1,\,q]}$ of the form:
\begin{align*}
\mathbf{x}_i = \alpha_i \, \mathbf{t}_i + (1 - \alpha_i) \, \mathbf{b}_i, ~~~~0<\alpha_i \leq 1\,,
\end{align*}
where $\mathbf{t}_i$ represents the known target that replaces a fraction $\alpha_i$ of the background $\mathbf{b}_i$ (i.e. at the same spatial location). The remaining ($e-q$) pixels in the given HSI, with $e = h \times w$, are thus only background ($\alpha=0$).
$\\$
\item We assume that all $\{\mathbf{t}_i\}_{i\in[1,\,q]}$ consist of similar materials, thus they should be represented by a linear combination of $N_t$ common target samples $\{\mathbf{a}^t_{j}\}_{j \in [1, \, N_t]}$, where $\mathbf{a}_j^t \in\mathbb{R}^p$ (the superscript $t$ is for target), but weighted with different set of coefficients $\{\beta_{i,j}\}_{j\in[1, N_t]}$. Thus, each of the $q$ targets is represented as:
\begin{align*}
\mathbf{x}_i = \alpha_i \sum\limits_{j=1}^{N_t} \Big(\beta_{i,j} \, \mathbf{a}^t_{j} \Big) + (1 - \alpha_i) \, \mathbf{b}_i \hspace{0.5cm} i \in [1,q]\, .
\end{align*}
\item We rearrange the given HSI into a two-dimensional matrix $\mathbf{D}\in \mathbb{R}^{e \times p}$, with $e = h \times w$ (by lexicographically ordering the columns). This matrix $\mathbf{D}$, can be decomposed into a low rank matrix $\mathbf{L}_0$ representing the pure background, a sparse matrix capturing any spatially small signal residing in the known target subspace, and a noise matrix $\mathbf{N}_0$. More precisely, the model is:
\begin{align*}\label{eq:mod2}
\small
\mathbf{D} &= \mathbf{L}_0 + (\mathbf{A}_t \, \mathbf{C}_0)^T + \mathbf{N}_0
\end{align*}
where $(\mathbf{A}_t\mathbf{C}_0)^T$ is the sparse target matrix, ideally with $q$ non-zero rows representing $\alpha_i \{\mathbf{t}^T_i\}_{i\in[1,q]}$ , with target dictionary $\mathbf{A}_t \in \mathbb{R}^{p \times N_t}$ having columns representing target samples $\{\mathbf{a}^t_{j}\}_{j \in [1, N_t]}$, and a coefficient matrix $\mathbf{C}_0\in\mathbb{R}^{N_t \times e}$ that should be a sparse column matrix, again ideally containing $q$ non-zero columns each representing $\alpha_i [\beta_{i,1}, \, \cdots, \, \beta_{i, N_t}]^T$, $i \in [1, q]$. $\mathbf{N}_0$ is assumed to be independent and identically distributed Gaussian noise with zero mean and unknown standard deviation.
$\\$
\item After reshaping $\mathbf{L}_0$, $(\mathbf{A}_t\, \mathbf{C}_0)^T$ and $\mathbf{N}_0$ back to a cube of size $h \times w \times p$, we call these entities the ``low rank background HSI'', ``sparse target HSI'', and ``noise HSI'', respectively.
\end{enumerate}
In order to recover the low rank matrix $\mathbf{L}_0$ and sparse target matrix $(\mathbf{A}_t\mathbf{C}_0)^T$, we consider the following minimization problem:
\vspace{-0.8mm}
\begin{equation}
\label{eq:ert5}
\underset{\mathbf{L}, \mathbf{C}} {\mathrm{min}} \,  \Bigl\{\tau \, \mathrm{rank}(\mathbf{L})+ \lambda \, \left\Vert\mathbf{C}\right\Vert_{2,0} +  \left\Vert\mathbf{D} - \mathbf{L} - (\mathbf{A}_t\mathbf{C})^T\right\Vert_F^2 \Bigr\}\,,
\end{equation}
where $\tau$ controls the rank of $\mathbf{L}$, and $\lambda$ the sparsity level in $\mathbf{C}$.

\subsection{Recovering a low rank background matrix and a sparse target matrix by convex optimization}
We relax the rank term and the $||.||_{2,0}$ term to their convex proxies. More precisely, we use nuclear norm $||\mathbf{L}||_*$  as a surrogate for the rank$(\mathbf{L})$ term, and the $l_{2,1}$ norm for the $l_{2,0}$ norm\footnote{A natural suggestion could be that the rank of $\mathbf{L}$ usually has a physical meaning (e.g., number of endmembers in background), and thus, why not to minimize the latter two terms in eq. \eqref{eq:convex_model} with the constraint that the rank of $\mathbf{L}$ should not be larger than a fixed value $d$? That is,
\begin{center}
$\underset{\mathbf{L}, \mathbf{C}} {\mathrm{min}} \, \Bigl\{\lambda \,||\mathbf{C}||_{2,1} +  ||\mathbf{D} - \mathbf{L} - (\mathbf{A}_t \mathbf{C})^T||_F^2 \Bigr\}\,,
~~s.t.~~ \mathrm{rank}(\mathbf{L}) \leq d$.
\end{center}
In our opinion, assuming that the number of endmembers in background is known exactly will be a strong assumption and our work will be less general as a result. One can assume $d$ to be some upper bound, in which case, the suggested formulation is a possible one. However, solving such a problem (with a hard constraint that the rank should not exceed some bound) is in general a NP-hard problem, unless there happens to be some special form in the objective which allows for a tractable solution. Thus, we adopt the soft constraint form with the nuclear norm as a proxy for the rank of $\mathbf{L}$; this is an approximation commonly done in the field and is found to give good solutions in many problems empirically.}.
\\
\\
We now need to solve the following convex minimization problem:
\vspace{-0.8mm}
\begin{align}{\label{eq:convex_model}}
\small
\underset{\mathbf{L}, \mathbf{C}} {\mathrm{min}} \, \Bigl\{\tau \,\left\Vert\mathbf{L}\right\Vert_*+ \lambda \,\left\Vert\mathbf{C}\right\Vert_{2,1} +  \left\Vert\mathbf{D} - \mathbf{L} - (\mathbf{A}_t \mathbf{C})^T\right\Vert_F^2 \Bigr\}\,,
\end{align}
Problem \eqref{eq:convex_model} is solved via an alternating minimization of two subproblems. Specifically, at each iteration $k$:
\begin{subequations}\label{eq:sub}
\footnotesize
\begin{alignat}{2}
\label{eq:sub1a}
\mathbf{L}^{(k)} &= \underset{\mathbf{L}} {\mathrm{argmin}} \, \Bigl\{\left\Vert\mathbf{L} - \left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T\right)\right\Vert_F^2 + \tau \,\left\Vert\mathbf{L}\right\Vert_* \, \Bigr\}\,, \\
\label{eq:sub2b}
\mathbf{C}^{(k)} &= \underset{\mathbf{C}} {\mathrm{argmin}} \, \Bigl\{\left\Vert\left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T -  \mathbf{A}_t \,  \mathbf{C}\right\Vert_F^2 + \lambda \,\left\Vert\mathbf{C}\right\Vert_{2,1} \, \Bigr\}.
\end{alignat}
\end{subequations}
The minimization sub-problems \eqref{eq:sub1a} \eqref{eq:sub2b} are convex and each can be solved optimally.
\\
\\
{\bf Solving sub-problem \eqref{eq:sub1a}:}
We solve subproblem \eqref{eq:sub1a} via the Singular Value Thresholding operator \cite{SVT2010}. We assume that $\left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T \right)$ has a rank equal to $r$.
According to theorem 2.1 in \cite{SVT2010}, subproblem \eqref{eq:sub1a} admits the following closed-form solution:
\begin{center}
\noindent \fbox{\parbox{7cm}{%
$~~~~~~~~~~~~\mathbf{L}^{(k)} = D_{\tau/2}\left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T \right)$\vspace{0.3cm}\\
$~~~~~~~~~~~~~~~~~~~~= \mathbf{U}^{(k)} \, D_{\tau/2} \left( \mathbf{S}^{(k)} \right) \, \mathbf{V}^{(k)T}$\vspace{0.3cm}\\
$~~~~~~~~~~~~~~~~~~~~= \mathbf{U}^{(k)} \, \diag \left(\left\{ \left(s_t^{(k)} - \frac{\tau}{2}\right)_+ \right\} \right) \, \mathbf{V}^{(k)T}$
}}
\end{center}

where $\mathbf{S}^{(k)} = \diag \left( \left\{s_t^{(k)}\right\}_{1 \leq t \leq r}\right)$, and $D_{\tau/2}(.)$ is the singular value shrinkage operator.
\\
The matrices $\mathbf{U}^{(k)} \in \mathbb{R}^{e \times r}$, $\mathbf{S}^{(k)} \in \mathbb{R}^{r \times r}$ and $\mathbf{V}^{(k)} \in \mathbb{R}^{p \times r}$ are generated by the singular value decomposition of $\left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T \right)$.

\begin{proof}
Since the function $\Bigl\{\left\Vert\mathbf{L} - \left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T\right)\right\Vert_F^2 + \tau \,\left\Vert\mathbf{L}\right\Vert_* \, \Bigr\}$ is strictly convex, it is easy to see that there exists a unique minimizer, and we thus need to prove that it is equal to $D_{\tau/2}\left(\mathbf{D} - \left(\mathbf{A}_t\, \mathbf{C}^{(k-1)}\right)^T \right)$. Note that to understand about how the aforementioned closed-form solution has been obtained, we provide in detail the proof steps that have been given in \cite{SVT2010}.
\\
To do this, let us first find the derivative of subproblem \eqref{eq:sub1a} w.r.t. $\mathbf{L}$ and set it to zero. We obtain:
\begin{equation}
\label{eq:r45t}
\left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T \right) - \hat{\mathbf{L}} = \frac{\tau}{2} \, \partial \left\Vert\hat{\mathbf{L}}\right\Vert_*
\end{equation}
Set $\hat{\mathbf{L}} = D_{\tau/2}\left(\mathbf{D} - \left(\mathbf{A}_t \,\mathbf{C}^{(k-1)}\right)^T \right)$ for short. In order to show that $\hat{\mathbf{L}}$ obeys \eqref{eq:r45t}, suppose the singular value decomposition (SVD) of $\left(\mathbf{D} - \left(\mathbf{A}_t \,\mathbf{C}^{(k-1)}\right)^T \right)$ is given by:
\begin{equation*}
\left(\mathbf{D} - \left(\mathbf{A}_t \,\mathbf{C}^{(k-1)}\right)^T \right) = \mathbf{U}_0 \, \mathbf{S}_0\,\mathbf{V}_0^T + \mathbf{U}_1\,\mathbf{S}_1\,\mathbf{V}_1^T \, ,
\end{equation*}
where $\mathbf{U}_0$, $\mathbf{V}_0$ (resp. $\mathbf{U}_1$, $\mathbf{V}_1$) are the singular vectors associated with singular values larger than $\tau/2$ (resp. inferior than or equal to $\tau/2$).
With these notations, we have:
\begin{equation*}
\hat{\mathbf{L}} =  D_{\tau/2}\left(\mathbf{U}_0 \,\mathbf{S}_0 \,\mathbf{V}_0^T \right) = \left(\mathbf{U}_0 \, \left(\mathbf{S}_0 - \frac{\tau}{2}\, \mathbf{I}\right)\, \mathbf{V}_0^T \right)
\end{equation*}
Thus, if we return back to equation \eqref{eq:r45t}, we obtain:
\begin{align*}
\mathbf{U}_0 \, \mathbf{S}_0\,\mathbf{V}_0^T + \mathbf{U}_1\,\mathbf{S}_1\,\mathbf{V}_1^T - \mathbf{U}_0\, \left(\mathbf{S}_0 - \frac{\tau}{2}\, \mathbf{I}\right)\,\mathbf{V}_0^T = \frac{\tau}{2} \, \partial \left\Vert\hat{\mathbf{L}}\right\Vert_* \, ,
\\
\Rightarrow \mathbf{U}_1\, \mathbf{S}_1\,\mathbf{V}_1^T + \mathbf{U}_0 \,\frac{\tau}{2}\,\mathbf{V}_0^T = \frac{\tau}{2} \, \partial \left\Vert\hat{\mathbf{L}}\right\Vert_* \, ,
\\
\Rightarrow \left(\mathbf{U}_0\, \mathbf{V}_0^T + \mathbf{W}\right) = \partial \left\Vert\hat{\mathbf{L}}\right\Vert_* \, ,
\end{align*}
where $\mathbf{W} = \displaystyle \frac{2}{\tau} \, \mathbf{U}_1\, \mathbf{S}_1\, \mathbf{V}_1^T$. \\

Let $\mathbf{U}_L\, \mathbf{S}_L\, \mathbf{V}_L$ be the SVD of $\mathbf{L}$, it is known \cite{4797640, Lewis2003, WATSON199233} that
\begin{equation*}
\partial \left\Vert\hat{\mathbf{L}}\right\Vert_* = \left\{ \mathbf{U}_L\, \mathbf{V}_L^T + \mathbf{W} \, : \, \mathbf{W}\in\mathbb{R}^{e \times p}, \, \mathbf{U}_L^T\, \mathbf{W} = \mathbf{0}, \, \mathbf{W}\, \mathbf{V}_L = \mathbf{0}, \, \left\Vert\mathbf{W}\right\Vert_2 \leq 1\right\} \, .
\end{equation*}
Hence, $\left(\mathbf{D} - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T \right) - \hat{\mathbf{L}} = \displaystyle\frac{\tau}{2} \, \partial \left\Vert\hat{\mathbf{L}}\right\Vert_*$, which concludes the proof.
\end{proof}
$\\$
{\bf Solving sub-problem \eqref{eq:sub2b}:}
\eqref{eq:sub2b} can be solved by various methods, among which we adopt the Alternating Direction Method of Multipliers (ADMM) \cite{Boyd:2011:DOS:2185815.2185816}.
More precisely, we introduce an auxiliary variable $\mathbf{F}$ into sub-problem \eqref{eq:sub2b} and recast it into the following form:
\begin{equation}
\label{eq:problemm}
\left(\mathbf{C}^{(k)}, \mathbf{F}^{(k)}\right) = \underset{s.t.~~ \mathbf{C} = \mathbf{F}} {\mathrm{argmin}} \, \left\{\left\Vert \left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T - \mathbf{A}_t \, \mathbf{C}\right\Vert_F^2 + \lambda \, \left\Vert\mathbf{F}\right\Vert_{2,1}\right\}
\end{equation}
Problem \eqref{eq:problemm} is then solved as follows (scaled form of ADMM):
\begin{subequations}\label{eq:subbb}
\small
\begin{alignat}{2}
\label{eq:sub11a}
\mathbf{C}^{(k)} &= \underset{\mathbf{C}} {\mathrm{argmin}} \, \left\{\left\Vert \left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T - \mathbf{A}_t \,\mathbf{C}\right\Vert_F^2 + \frac{\rho^{(k-1)}}{2}\, \left\Vert\mathbf{C} - \mathbf{F}^{(k-1)} + \frac{1}{\rho^{(k-1)}}\, \mathbf{Z}^{(k-1)}\right\Vert_F^2 \, \right\}\,, \\
\label{eq:sub22b}
\mathbf{F}^{(k)} &= \underset{\mathbf{F}} {\mathrm{argmin}} \, \left\{\lambda \, \left\Vert\mathbf{F}\right\Vert_{2,1} + \frac{\rho^{(k-1)}}{2}\,\left\Vert \mathbf{C}^{(k)} - \mathbf{F} + \frac{1}{\rho^{(k-1)}} \, \mathbf{Z}^{(k-1)}\right\Vert_F^2\right\}\,, \\
\label{eq:sub33c}
\mathbf{Z}^{(k)} &= \mathbf{Z}^{(k-1)} + \rho^{(k-1)} \,\left(\mathbf{C}^{(k)} - \mathbf{F}^{(k)}\right)\,,
\end{alignat}
\end{subequations}
where $\mathbf{Z} \in \mathbb{R}^{N_t \times e}$ is the Lagrangian multiplier matrix, and $\rho$ is a positive scalar.
\\
\\
{\em Solving sub-problem \eqref{eq:sub11a}}:
\begin{align*}
-2\, \mathbf{A}_t^T \left(\left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T - \mathbf{A}_t\, \mathbf{C} \right) + \rho^{(k-1)} \, \left( \mathbf{C} - \mathbf{F}^{(k-1)} + \frac{1}{\rho^{(k-1)}} \, \mathbf{Z}^{(k-1)}\right) = \mathbf{0} \, ,
\\
\Rightarrow \left(2\, \mathbf{A}_t^T \, \mathbf{A}_t + \rho^{(k-1)}\, \mathbf{I} \right) \, \mathbf{C} = \rho^{(k-1)} \, \mathbf{F}^{(k-1)} - \mathbf{Z}^{(k-1)} + 2\, \mathbf{A}_t^T \, \left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T \, .
\end{align*}
This implies:
\[
 \boxed{\mathbf{C}^{(k)} = \left( 2\, \mathbf{A}_t^T \, \mathbf{A}_t + \rho^{(k-1)} \, \mathbf{I}\right)^{-1} \, \left(\rho^{(k-1)}\, \mathbf{F}^{(k-1)} - \mathbf{Z}^{(k-1)} + 2\, \mathbf{A}_t^T \, \left(\mathbf{D} - \mathbf{L}^{(k)}\right)^T \right)} \, .
 \]
\\
{\em Solving sub-problem \eqref{eq:sub22b}}:
\\
According to Lemma 3.3 in \cite{doi:10.1137/080730421} and Lemma 4.1 in \cite{6180173}, subproblem \eqref{eq:sub22b} admits the following closed form solution:
\[
 \boxed{[\mathbf{F}]_{:, j}^{(k)}  = \max \left( \left\Vert\left[\mathbf{C}\right]_{:,j}^{(k)} + \frac{1}{\rho^{(k-1)}} \, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}\right\Vert_2 - \frac{\lambda}{\rho^{(k-1)}}, \, 0 \right) \, \left( \displaystyle \frac{\left[\mathbf{C}\right]_{:,j}^{(k)} +
 {\frac{1}{\rho^{(k-1)}}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}}{\left\Vert  \left[\mathbf{C}\right]_{:,j}^{(k)} + \frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}\right\Vert_2}\right)
 }
 \]

 \\
 \begin{proof}
 At the $j^{th}$ column, subproblem \eqref{eq:sub22b} refers to:
 \begin{equation*}
 \left[\mathbf{F}\right]_{:,j}^{(k)} = \underset{ \left[\mathbf{F}\right]_{:,j}} {\mathrm{argmin}} \, \left\{\lambda \, \left\Vert \left[\mathbf{F}\right]_{;,j}\right\Vert_2 + \frac{\rho^{(k-1)}}{2}\,\left\Vert\left[\mathbf{C}\right]_{:,j}^{(k)} - \left[\mathbf{F}\right]_{:,j} + \frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}\right\Vert_2^2 \, \right\} \, .
 \end{equation*}
 By finding the derivative w.r.t $[\mathbf{F}]_{:,j}$ and setting it to zero, we obtain:
\begin{eqnarray}
 \label{eq:huyrz}
 -\rho^{(k-1)} \, \left( \left[\mathbf{C}\right]_{:,j}^{(k)} -  \left[\mathbf{F}\right]_{:,j} + \frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)} \right) + \frac{\lambda \, \left[\mathbf{F}\right]_{:,j}}{\left\Vert\left[\mathbf{F}\right]_{:,j}\right\Vert_2} = \mathbf{0}\, \nonumber
 \\
 \Rightarrow \left[\mathbf{C}\right]_{:,j}^{(k)} + \frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)} = \left[\mathbf{F}\right]_{:,j} + \frac{\lambda \, \left[\mathbf{F}\right]_{:,j}}{\rho^{(k-1)}\, \left\Vert\left[\mathbf{F}\right]_{:,j}\right\Vert_2} \, .
\end{eqnarray}
 By computing the $l_2$ norm of \eqref{eq:huyrz}, we obtain:
 \begin{equation}
  \label{eq:bbcv}
 \left\Vert \left[\mathbf{C}\right]_{:,j}^{(k)} + \frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}\right\Vert_2 = \left\Vert\left[\mathbf{F}\right]_{:,j}\right\Vert_2 + \frac{\lambda}{\rho^{(k-1)}} \, .
 \end{equation}
From equation \eqref{eq:huyrz} and equation \eqref{eq:bbcv}, we have:
\begin{equation}
\label{eq:wwwe}
\displaystyle \frac{\left[\mathbf{C}\right]_{:,j}^{(k)} + \displaystyle\frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}}{ \left\Vert\left[\mathbf{C}\right]_{:,j}^{(k)} + \displaystyle\frac{1}{\rho^{(k-1)}}\, \left[\mathbf{Z}\right]_{:,j}^{(k-1)}\right\Vert_2} = \frac{\left[\mathbf{F}\right]_{:,j}}{\left\Vert\left[\mathbf{F}\right]_{:,j}\right\Vert_2} \, .
\end{equation}
Consider that:
\begin{equation}
\label{eq:heqrtqr}
[\mathbf{F}]_{:,j}=  \left\Vert[\mathbf{F}]_{:,j}\right\Vert_2 \times \displaystyle \frac{[\mathbf{F}]_{:,j}}{ \left\Vert[\mathbf{F}]_{:,j}\right\Vert_2} \, .
\end{equation}
By replacing $ \left\Vert[\mathbf{F}]_{:,j}\right\Vert_2$ from \eqref{eq:bbcv} into \eqref{eq:heqrtqr}, and $\displaystyle\frac{[\mathbf{F}]_{:,j}}{ \left\Vert[\mathbf{F}]_{:,j}\right\Vert_2}$ from \eqref{eq:wwwe} into \eqref{eq:heqrtqr}, we conclude the proof.
 \end{proof}

 \subsection{Some initializations and convergence criterion}
 We initialize $\mathbf{L}^{(0)} = \mathbf{C}^{(0)} = \mathbf{F}^{(0)} = \mathbf{Z}^{(0)} = \boldsymbol{0}$, $\rho^{(0)} = 10^{-4}$ and update $\rho^{(k)} = 1.1 \, \rho^{(k-1)}$. The criteria for convergence of sub-problem \eqref{eq:sub2b} is $ \left\Vert\mathbf{C}^{(k)} - \mathbf{F}^{(k)} \right\Vert_F^2   \leq 10^{-6}$.
 \\
 \\
For Problem \eqref{eq:convex_model}, we stop the iteration when the following convergence criterion is satisfied:
\begin{align*}
\frac{ \left\Vert\mathbf{L}^{(k)} - \mathbf{L}^{(k-1)} \right\Vert_F}{ \left\Vert\mathbf{D}\right\Vert_F} \leq \epsilon~~~~~\text{and} ~~~~~
\frac{ \left\Vert\left(\mathbf{A}_t \, \mathbf{C}^{(k)}\right)^T - \left(\mathbf{A}_t \, \mathbf{C}^{(k-1)}\right)^T\right\Vert_F}{ \left\Vert\mathbf{D}\right\Vert_F} \leq \epsilon
\end{align*}
where $\epsilon>0$ is a precision tolerance parameter. We set $\epsilon = 10^{-4}$.

\subsection{Our novel target detector: $(\mathbf{A}_t\mathbf{C})^T$}
\label{sec:first_dete_strat}
We use $(\mathbf{A}_t \mathbf{C})^T$ directly for the detection. Note that for this detector, we require as few false alarms as possible to be deposited in the target image, but we do not need the target fraction to be entirely removed from the background (that is, a very weak target separation can suffice). As long as enough of the target fractions are moved to the target image, such that non-zero support is detected at the corresponding pixel location, it will be adequate for our detection scheme. From this standpoint, we should choose a $\lambda$ value that is relatively large so that the target image is really sparse with zero or little false alarms, and only the signals that reside in the target subspace specified by $\mathbf{A}_t$ will be deposited there.

\section{Experiments and Analysis}
\label{sec:experimentss}
To obtain the same scene as in Fig. 8 in \cite{Swayze10245}, we have concatenated two sectors labeled as  ``f970619t01p02\_r02\_sc03.a.rf'' and ``f970619t01p02\_r02\_sc04.a.rfl'' from the online Cuprite data \cite{CupriteHSIOnline}. We shall call the resulting HSI as ``Cuprite HSI'' (see Fig. \ref{fig:cupriteHSI}). The Cuprite HSI is a mining district area, which is well understood mineralogically \cite{Swayze10245, JGRE:JGRE1642}. It contains well-exposed zones of advanced argillic alteration, consisting principally of kaolinite, alunite, and hydrothermal silica. It was acquired by the Airborne Visible/Infrared Imaging Spectrometer (AVIRIS) in June 23, 1995 at local noon and under high visibility conditions by a NASAER-2 aircraft flying at an altitude of 20 km. It is a $1024 \times 614$ image and consists of 224 spectral (color) bands in contiguous (of about 0.01 $\mu$m) wavelengths ranging exactly from 0.4046 to 2.4573 $\mu$m. Prior to some analysis of the Cuprite HSI, the spectral bands 1-4, 104-113, and 148-167 are removed due to the water absorption in those bands. As a result, a total of 186 bands are used.
 \begin{figure}[!tbp]
\centering
\minipage{2.5\textwidth}
  \hspace{-1.6cm}\includegraphics[width=0.5\linewidth]{ggh_new-eps-converted-to.pdf}
\endminipage
\caption{The Cuprite HSI of size $1024 \times 614 \times 186$. We exhibit the mean power in db over the 186 spectral bands.}\label{fig:cupriteHSI}
\end{figure}
 \begin{figure}[!tbp]
\minipage{0.50\textwidth}
  \includegraphics[width=\linewidth]{alunite_samples_new2-eps-converted-to.pdf}
\endminipage\hfill
\minipage{0.50\textwidth}
  \includegraphics[width=\linewidth]{original_image_alunite_new_cube.png}
\endminipage\hfill
\centering
\minipage{0.50\textwidth}
 \includegraphics[width=\linewidth]{original_image_alunite_new-eps-converted-to.pdf}
\endminipage
\caption{A $100 \times 100 \times 186$ ``Alunite HSI'' generated by 72 pure alunite samples picked from the Cuprite HSI (72 pixels from the solid red ellipses in Fig. \ref{fig:cupriteHSI}). For the third image, we exhibit the mean power in db over the 186 spectral bands.}\label{fig:image2}
\end{figure}
\begin{figure}[!]
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{alunite_new-eps-converted-to.pdf}
  \begin{center}
  \bf (a)
  \end{center}
\endminipage\hfill
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{kaolinite-eps-converted-to.pdf}
    \begin{center}
  \bf (b)
  \end{center}
\endminipage
\caption{Three-point band depth images for both {\bf(a)} alunite and {\bf(b)} kaolinite.}\label{fig:image3}
\end{figure}
$\\$
$\\$
By referring to Figure 8 in \cite{Swayze10245}, we picked 72 pure alunite pixels from the Cuprite HSI (72 pixels located inside the solid red ellipses in Fig. \ref{fig:cupriteHSI}) and generate a $100 \times 100 \times 186$ HSI zone formed by these pixels. We shall call this small HSI zone as ``Alunite HSI'' (see Fig. \ref{fig:image2}), and which will be used for the target evaluations later. We incorporate, in this zone, seven target blocks (each of size $6\times 3$) with $\alpha\in[0.01, \, 1]$ (all have the same $\alpha$ value), placed in long convoy formation all formed by the same target $\mathbf{t}$ that we picked from the Cuprite HSI and which will constitute our target of interest to be detected. The target $\mathbf{t}$ replaces a fraction $\alpha\in[0.01, \, 1]$ from the background; specifically, the following values of $\alpha$ are considered: 0.01, 0.02, 0.05, 0.1, 0.3, 0.5, 0.8, and 1.
\\
In the experiments, two kinds of targets $\mathbf{t}$ are considered:
\begin{enumerate}
\item `$\mathbf{t}$' that represents the buddingtonite target,
\item `$\mathbf{t}$' that represents the kaolinite target.
\end{enumerate}
More precisely, our detector $\left(\mathbf{A}_t\mathbf{C}\right)^T$   is evaluated on two target detection scenarios:
\begin{itemize}
\item{\bf Evaluation on an easy target (buddingtonite target):}  It has been noted by Gregg et. al. \cite{Swayze10245} that the ammonia in the Tectosilicate mineral type, known as buddingtonite, has a distinct N-H combination absorption at 2.12 $\mu$m, a position similar to that of the cellulose absorption in dried vegetation, from which it can be distinguished based on its narrower band width and asymmetry.  Hence, the buddingtonite mineral can be considered as an ``easy target''
because it does not look like any other mineral with its distinct 2.12$\mu$m absorption (that is, it is easily recognized based on its unique 2.12$\mu$m absorption band).
\\
In the experiments, we consider the ``buddingtonite'' pixel at location (731, 469) in the Cuprite HSI (the center of the dash-dotted yellow circle in Fig. \ref{fig:cupriteHSI}) as the buddingtonite target $\mathbf{t}$ to be incorporated in the Alunite HSI for $\alpha\in[0.01, \, 1]$.
$\\$
\item{\bf Evaluation on a challenging target (kaolinite target)}\footnote{We thank Dr. Gregg A. Swayze from the United States Geological Survey (USGS) who has suggested us to evaluate our model \eqref{eq:convex_model} on the distinction between alunite and kaolinite minerals.}{\bf :} The paradigm in military applications hyperspectral imagery seems to center on finding the target but ignoring all the rest. Sometimes, that rest is important, especially if the target is well matched to the surroundings. It has been shown by Gregg et al. \cite{Swayze10245} that both alunite and kaolinite minerals have overlapping spectral features, and thus, discrimination between these two minerals is very challenging \cite{doi:10.1029/2002JE001847, Swayze10245}.
\\
In the experiments, we consider the ``kaolinite'' pixel at location (672, 572) in the Cuprite HSI (the center of the dotted blue circle in Fig. \ref{fig:cupriteHSI}) as the kaolinite target $\mathbf{t}$ to be incorporated in the Alunite HSI for $\alpha\in[0.01, \, 1]$.
\\
Fig. 4(a) exhibits a three-point band depth image for our alunite background that shows the locations where an absorption feature, centered near 2.17 $\mu$m, is expressed in spectra of surface materials. Fig. 4(b) exhibits a three-point band depth image for our kaolinite target that shows the locations where an absorption feature, centered near 2.2 $\mu$m, is expressed in spectra of surface materials. As we can observe, there is subtle difference between the alunite and kaolinite three-point band depth images, showing that the successful spectral distinction between these two minerals is a very challenging task to achieve\footnote{We have been inspired by Fig. 8D-E in \cite{doi:10.1029/2002JE001847} to provide a close example of it in this chapter as can be shown in Fig. \ref{fig:image3}.} \cite{doi:10.1029/2002JE001847}.

\end{itemize}

\subsection{Construction of the target dictionary $\mathbf{A}_t$}
\label{sec:dict_const}
 \begin{figure}[!tbp]
  \begin{center}
\large \bf When $\mathbf{A}_t$ is constructed from background samples
\end{center}
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha1_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha1_poordictionary_buddingtonite.png}
\endminipage\hfill
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha1_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha1_poordictionary_kaolinite.png}
\endminipage
\caption{Evaluation of our detector $(\mathbf{A}_t\mathbf{C})^T$ for detecting the buddingtonite and kaolinite target (for $\alpha=1$) from the Alunite HSI when $\mathbf{A}_t$ is contsructed from some background pixels acquired from the Alunite HSI.} \label{fig:poor_dictionary}
\end{figure}
An important problem that requires a very careful attention is the construction of an appropriate dictionary $\mathbf{A}_t$ in order to well capture the target and distinguish it from the background. Thus, if $\mathbf{A}_t$ does not well represent the target of interest, our model in \eqref{eq:convex_model} may fail on discriminating the targets from the background. For example, Fig. \ref{fig:poor_dictionary} shows the detection results of our detector $(\mathbf{A}_t\mathbf{C})^T$ when $\mathbf{A}_t$ is constructed from some of the background pixels in the Alunite HSI. We can obviously observe that our detector is not able to capture the targets mainly because of the poor dictionary $\mathbf{A}_t$ constructed.
\\
In reality, the target present in the HSI can be highly affected by the atmospheric conditions, sensor noise, material composition, and scene geometry, that may produce huge variations on the target spectra. In view of these real effects, it is very difficult to model the target dictionary $\mathbf{A}_t$ well. But this raises the question on ``{\em how these effects should be dealt with?}''.
\\
Some scenarios for modelling the target dictionary have been suggested in the literature. For example, by using physical models and the MORTRAN atmospheric modeling program \cite{Berk89}, target spectral signatures can be generated under various atmospheric conditions. For simplicity, we handle this problem in this work by exploiting target samples available in some online spectral libraries. More precisely, $\mathbf{A}_t$ can be constructed via the United States Geological Survey (USGS - Reston) spectral library \cite{Clark93}. However, the user can also deal with the Advanced Spaceborne Thermal Emission and Reflection (ASTER) spectral library \cite{Baldridge09} that includes data from the USGS spectral library, the Johns Hopkins University (JHU) spectral library, and the Jet Propulsion Laboratory (JPL) spectral library.
\\
\\
There are three buddingtonite samples available in the ASTER spectral library and will be considered to construct the dictionary $\mathbf{A}_t$ for the detection of our buddingtonite target (see Fig. \ref{fig:dictionaries_kao_bud} (first column)); whereas six kaolinite samples are available in the USGS spectral library and will be acquired to construct $\mathbf{A}_t$ for the detection of our kaolinite target (see Fig. \ref{fig:dictionaries_kao_bud} (second column)).
\\
\\
For instance, it is usually difficult to find, for a specific given target, a sufficient number of available samples in the online spectral libraries. Hence, the dictionary $\mathbf{A}_t$ can still not be sufficiently selective and accurate. This is the most reason why problem \eqref{eq:convex_model} may fail to well capture the targets from the background.

 \begin{figure}[!tbp]
 \minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{dictionary_buddingtonite-eps-converted-to.pdf}
\endminipage\hfill
\minipage{0.52\textwidth}
  \includegraphics[width=\linewidth]{dictionary_kaolinite-eps-converted-to.pdf}
\endminipage
\caption{Both target dictionaries for the detection of buddingtonite and kaolinite.}\label{fig:dictionaries_kao_bud}
\end{figure}

\newpage
 \subsection{Target Detection Evaluation}
We now aim to qualitatively evaluate the target detection performances of our detector $(\mathbf{A}_t\mathbf{C})^T$ on both buddingtonite and kaolinite target detection scenarios when $\mathbf{A}_t$ is constructed from target samples available in the online spectral libraries (from Fig. \ref{fig:dictionaries_kao_bud}). As can be seen from Fig. \ref{fig:detection_evaluations}, our detector is able to detect the buddingtonite targets with no false alarms until $\alpha \leq 0.1$ where a lot of false alarms appear.
\\
For the detection of kaolinite, it was difficult to have a clean detection (that is, without false alarms) especially for $0.1<\alpha\leq1$. This is to be expected since the kaolinite target is well matched to the alunite background (that is, kaolinite and alunite have overlapping spectral features), and hence, discrimination between them is very challenging.
\\
\\
It is interesting to note (results are omitted in this chapter) that if we consider $\mathbf{A}_t = \mathbf{t}$ (that is, we are searching for the exact signature $\mathbf{t}$ in the Alunite HSI), the buddingtonite and even the kaolinite targets are able to be detected with no false alarms for $0.1<\alpha\leq1$. When $\alpha\leq0.1$, a lot of false alarms appear, but, the detection performance for both buddngtonite and kaolinite targets remains better than to those in Fig. \ref{fig:detection_evaluations}.

 \begin{figure}[!tbp]
   \begin{center}
\large \bf When $\mathbf{A}_t$ is constructed from target samples
\end{center}
 \begin{center}
\large$\bf \boldsymbol{\alpha} = 1$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha1_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha1_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha1_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha1_onlinedictionary_kaolinite.png}
\endminipage

 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.8$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha08_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha08_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha08_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha08_onlinedictionary_kaolinite.png}
\endminipage

\end{figure}

 \begin{figure}[!tbp]
 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.5$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha05_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha05_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha05_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha05_onlinedictionary_kaolinite.png}
\endminipage

 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.3$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha03_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha03_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha03_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha03_onlinedictionary_kaolinite.png}
\endminipage

\end{figure}

 \begin{figure}[!tbp]
  \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.1$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha01_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha01_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha01_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha01_onlinedictionary_kaolinite.png}
\endminipage

 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.05$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha005_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha005_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha005_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha005_onlinedictionary_kaolinite.png}
\endminipage

\end{figure}

 \begin{figure}[!tbp]
 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.02$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha002_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha002_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha002_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha002_onlinedictionary_kaolinite.png}
\endminipage

 \begin{center}
\large$\bf \boldsymbol{\alpha} = 0.01$
\end{center}
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha001_buddingtonite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha001_onlinedictionary_buddingtonite_cube_new.png}
\endminipage\hfill

\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{originalHSI_alpha001_kaolinite_cube.png}
\endminipage\hfill
\minipage{0.54\textwidth}
  \includegraphics[width=\linewidth]{Ourdetector_alpha001_onlinedictionary_kaolinite.png}
\endminipage

\caption{Evaluation of our detector $(\mathbf{A}_t\mathbf{C})^T$ for detecting the buddingtonite and kaolinite target (for $\alpha \in [0.01, \, 1]$) when $\mathbf{A}_t$ is contsructed from target samples in the online spectral libraries.}\label{fig:detection_evaluations}
\end{figure}

\newpage

\section{Conclusion and Future Work}
 In this chapter, the well-known Robust Principal Component Analysis (RPCA) is exploited for target detection in hyperspectral imagery. By taking similar assumptions to those used in RPCA, a given hyperspectral image (HSI) has been decomposed into the sum of a low rank background HSI (denoted as $\mathbf{L}$) and a sparse target HSI (denoted as $\mathbf{E}$) that only contains the targets (with the background is suppressed) \cite{8677268}. In order to alleviate the inadequacy of RPCA on distinguishing the true targets from the background, we have incorporated into the RPCA imaging, the prior target information that can often be provided to the user. In this regard, we have constructed a pre-learned target dictionary $\mathbf{A}_t$, and thus, the given HSI is being decomposed as the sum of a low rank background HSI $\mathbf{L}$ and a sparse target HSI denoted as $\left(\mathbf{A}_t\mathbf{C}\right)^T$, where $\mathbf{C}$ is a sparse activation matrix.
 \\
In this work, the sparse component $\left(\mathbf{A}_t\mathbf{C}\right)^T$ was only the object of interest, and thus, used directly for the detection. More precisely, the targets are deemed to be present at the non-zero entries of the sparse target HSI. Hence, a novel target detector is developed and which is simply a sparse HSI generated automatically from the original HSI, but containing only the targets of interests with the background is suppressed.
 \\
The detector is evaluated on real experiments, and the results of which demonstrate its effectiveness for hyperspectral target detection, especially on detecting targets that have overlapping spectral features with the background.
\\
\\
The $l_1$-norm regularizer, a continuous and convex surrogate, has been studied extensively in the literature \cite{Tibshirani96, 10.2307/3448465} and has been applied successfully to many applications including signal/image processing, biomedical informatics, and computer vision \cite{doi:10.1093/bioinformatics/btg308, 4483511, Beck:2009:FIS:1658360.1658364, 4799134, Ye:2012:SMB:2408736.2408739}. Although the $l_1$ norm based sparse learning formulations have achieved great success, they have been shown to be suboptimal in many cases \cite{CandAs2008, Zhanggg, zhang2013}, since the $l_1$ is still too far away from the ideal $l_0$ norm. To address this issue, many non-convex regularizers, interpolated between the $l_0$ norm and the $l_1$ norm, have been proposed to better approximate
the $l_0$ norm.  They include $l_q$ norm ($0<q<1$) \cite{Fourc1624}, Smoothly Clipped Absolute Deviation \cite{Fan01}, Log-Sum Penalty \cite{Candes08}, Minimax Concave Penalty \cite{zhang2010aa}, Geman Penalty \cite{392335, 5153291}, and Capped-$l_1$ penalty \cite{Zhanggg, zhang2013, Gong12}.
\\
In this regard, from problem \eqref{eq:convex_model}, it will be interesting to use other proxies than the $l_{2,1}$ norm, closer to $l_{2,0}$, in order to probably alleviate the $l_{2,1}$ artifact and also the manual selection problem of both $\tau$ and $\lambda$.
But although the non-convex regularizers (penalties) are appealing in sparse learning, it remains a very big challenge to solve the corresponding non-convex optimization problems.

\begin{acknowledgement}
The authors would greatly thank Dr.~Gregg A. Swayze from the United States Geological Survey (USGS) for his time in providing them helpful remarks about the cuprite data and especially on the buddingtonite, alunite and kaolinite minerals.
\end{acknowledgement}

\bibliographystyle{IEEEbib}


\section{Introduction}

Throught this article, we fix  a differential field $k$ of characteristic zero.  Let $E$ be a differential field extension of $k$ and let $'$ denote the derivation on $E$. We say that  $E$ is a \textit{ liouvillian extension} of $k$  if $E=k(t_1,\cdots,t_n)$ and there is a tower of differential fields
\begin{equation*}k=k_0\subset k_1\subset\cdots\subset k_n=E\end{equation*}
such that for each $i$,  $k_i=k_{i-1}(t_i)$ and either $t'_i\in k_{i-1}$ or $t'_i/t_i\in k_{i-1}$ or $t_i$ is
algebraic over $k_{i-1}$. If $E=k(t_1,\cdots,t_n)$ is a liouvillian extension of $k$ such that $t'_i\in k_{i-1}$ for each $i$ then we call $E$ an \textit{iterated antiderivative extension} of $k$. A solution of a differential equation over $k$  is said to be  liouvillian over $k$ if the solution belongs to  some liouvillian extension of $k$.

Let $\mathcal{P}$ be an $n+1$ variable polynomial over $k$. We are concerned with the  liouvillian solutions of the differential equation \begin{equation}\label{firstdiffereqn}\mathcal{P}(y,y',\cdots, y^{(n)})=0.\end{equation}   We prove in theorem \ref{solutiondense} that \begin{itemize}
\item[] if $E$ is a liouvillian extension of $k$ and  $K$ is a differential field intermediate to $E$ and $k$ then $K=k\langle u_1,\cdots,u_l\rangle$ where for each $i$, the element $u_i$ satisfies a linear  differential equation over $k\langle u_1,\cdots, u_{i-1}\rangle$. Moreover if $E$ is an iterated antiderivative extension of $k$ having the same field of constants as $k$ then each $u_i$ can be chosen so that $u'_i\in k(u_1,\cdots, u_{i-1})$, that is,   $K$ is also an iterated antiderivative extension of $k$. \end{itemize}

Our result regarding iterated antiderivative extensions generalises the main result of \cite{Sri2010} to differential fields $k$ having a non algebraically closed  field of constants. The main ingredient used in the proof our theorem is the lemma \ref{ldesol} using which we also obtain the following interesting results concerning solutions of non linear differential equations.
\begin{itemize}

\item[A.] In remark \ref{order2}, we show that if $E$ is an iterated antiderivative extension of $k$ having the same field of constants and if $y\in E$ and $y\notin k$ satisfies a differential equation $y'=\mathcal{P}(y)$, where $\mathcal{P}$ is a polynomial in one variable over $k$, then degree of $\mathcal{P}$ must be less than or equal to $2$. \\
\item[B.] Let $C$ be an algebraically closed field of characteristic zero with the trivial derivation.  In proposition \ref{M-B}, we prove that for any rational function $\mathcal{R}$ in one variable over $C$, the differential equation $y'=\mathcal{R}(y)$ has a non constant liouvillian solution $y$  if and only if $1/\mathcal{R}(y)$ is of the form $\partial z/\partial y$ or $(1/az)(\partial z/ \partial y)$ for some $z\in C(y)$ and for some  non zero element $a\in C$.   This result generalises and provides a new proof for a result of Singer (see \cite{Sin1975}, corollary 2). We also prove that
 for any polynomial $\mathcal{P}$ in one variable over $C$ such that the degree of $\mathcal{P}$ is greater than or equal to $3$ and that $\mathcal{P}$ has no repeated roots,   the differential equation $(y')^2=\mathcal{P}(y)$ has no non constant liouvillian solution over $C$. This result appears as proposition \ref{hyperellipticcurves} and it generalises an observation made by Rosenlicht \cite{M.Ros} concerning non constant liouvillian solutions of the elliptic equation $(y')^2=y^3+ay+b$ over complex numbers with a non zero discriminant. \\
 \item[C.]  Using theorem \ref{solutiondense} one can construct a  family of differential equations with only algebraic solutions: Let $\a_2,\a_3,$ $\cdots,\a_n\in k$ such that $x'\neq \a_2$ and $x'\neq \a_3$ for any $x\in k$. Let $\overline{k}$ be  an algebraic closure of $k$ and let  $E$ be a  liouvillian extension  of $\overline{k}$ with $C_E=C_{\overline{k}}$. We prove in proposition \ref{algsolutions} that if there is an element $y\in E$  such that  $$y'=\a_ny^n+\cdots+\a_3y^3+\a_2y^2$$ then    $y\in \overline{k}$.

\end{itemize}

In a future publication, the author hopes to develop the techniques in this paper further to provide an algorithm to solve the following problem, which appears as ``Problem 7" in \cite{Sin1990}: Give a procedure to decide if a polynomial first order differential equation $\mathcal{P}(y,y')=0$, over the ordinary differential field $\mathbb{C}(x)$ with the usual derivation $d/dx$,  has an elementary solution and to find one if it does.

{\bf Preliminaries and Notations.}
 A \textsl{derivation} of the field $k$, denoted by $'$, is an additive endomorphism of $k$ that satisfies the Leibniz law $(xy)'=x'y+xy'$ for every $x,y\in k$. A field equipped
with a derivation map is called a \textsl{differential field}. For any $y\in k$, we will denote first and second derivative of $y$ by $y'$ and $y''$ respectively and for $n\geq 3$, the $n$th derivative of $y$ will be denoted by $y^{(n)}$. The set of \textsl{constants} $C_E$ of a differential field $E$ is the kernel of the endomorphism $'$ and it can be seen that the set of constants is a differential subfield of $k$.  Let $E$ and $k$  be differential fields. We say that $E$ is a \textsl{differential field extension}  of $k$ if $E$ is a field extension of $k$ and the restriction of the derivation of $E$ to $k$ coincides with the derivation of $k$. Whenever we write $k\subset E$ be differential fields, we mean that $E$ is a differential field extension of $k$ and we write $y\in E-k$ to mean that $y\in E$ and  $y\notin k$. The transcendence degree of a field extension $E$ of $k$ will be denoted by tr.d$(E|k)$. If $E$ is a field (respectively a differential field),  $M$ is a subfield (respectively a differential subfield) of $E$ and $K$ is a subset of $E$  then  the smallest subfield (respectively the smallest differential subfield) of $E$ containing both  $M$ and $K$ will be denoted by $M(K)$ (respectively by $M\langle K\rangle$). It is easy to see that the field $M(K)$ is a differential field if both $M$ and $K$ are differential subfields of a differential field.   It is well known that every derivation of $k$ can be uniquely extended to a derivation of any algebraic extension of $k$ and in particular, to any algebraic closure $\overline{k}$ of $k$. Thus if $E$ is a differential field extension of $k$ and tr.d$(E|k)=1$ then for any $y\in E$ transcendental over $k$ the derivation $\partial/\partial y$ on $k(y)$  uniquely extends to a derivation of  $E$.   We refer the reader to \cite{Kap}, \cite{Mag} and \cite{Put-Sin} for basic theory of differential fields and for the reader's convenience and for easy reference, we record a basic result concerning liouvillian extensions  in the following theorem.
\begin{theorem}\label{construction} Let $k$ be a differential field,  $E$ be a field extension of $k$ and $w\in E$ be transcendental over $k$. For an element $\a\in k$ if there is no element $x\in k$ such that $x'=\a$ then there is a unique derivation on $k(w)$ such that $w'=\a$ and that $C_{k(w)}=C_k$.  Similarly, if there is no element $x\in k$ such that $x'=l\a x$ for any positive integer $l$ then there is a unique derivation on $k(w)$ such that $w'=\a w$ and that $C_{k(w)}=C_k$. Finally, if $E$ is a differential field extension of $k$ having an element $u$ algebraic over $k$ such that $u'\in k$ then there is an element $\b\in k$ such that $u'=\b'$.   \end{theorem}

\section{Role of First Order Equations}

In this section we prove our main result.  The proof is based on the theory of linearly disjoint fields  and in particular, we will heavily rely on the following  lemma.

\lemma \label{ldesol}
Let $k\subset E$ be differential fields and let $K$ and $M$ be  differential fields  intermediate to  $E$ and $k$ such that $K$ and $M$ are linearly disjoint over $k$ as fields. Suppose that there is an element $y\in M(K)-M$ such that $y'=fy+g$ for some $f,g\in M$. Then there exist a monic linear  differential polynomial $L(Y)$ over $k$  of degree $\geq 1$ and an element $u\in K-k$ such that $L(u)=0$.

\begin{proof}

Let $\mathcal{B}:=\{e_\a\ |\ \a\in J\}$ be a  basis of the $k-$vector space $M$. Since $M$ and $K$ are linearly disjoint,  $\mathcal{B}$ is also a basis of $M(K)=K(M)$ as a $K-$vector space.   There exist elements  $\a_{1},\cdots,\a_r,$ $\a_{r+1},\cdots,\a_t\in J$ such that
\begin{equation}
y=\sum^r_{j=1}u_je_{\a_j},
\end{equation}
where $u_j\in K$ and for $i=1,\cdots, r,$
\begin{equation}
e'_{\a_i}=\sum^t_{p=1}n_{pi}e_{\a_p},\quad fe_{\a_i}=\sum^t_{p=1}l_{pi}e_{\a_p},\quad g=\sum^t_{p=1}m_pe_{\a_p},
\end{equation}
 where each $n_{pi}, l_{pi}$ and $m_p$ belongs to $k$.

Then \begin{align*}
y'&=\sum^r_{j=1}u'_je_{\a_j}+\sum^r_{j=1}u_j\sum^t_{p=1}n_{pj}e_{\a_p}\\
&=\sum^r_{j=1}u'_je_{\a_j}+\sum^t_{p=1}\left(\sum^r_{j=1}n_{pj}u_j\right)e_{\a_p}\\
&=\sum^r_{p=1}\left(u'_p+\sum^r_{j=1}n_{pj}u_j\right)e_{\a_p}+\sum^t_{p=r+1}\left(\sum^r_{j=1}n_{pj}u_j\right)e_{\a_p}
\end{align*}
and $fy=\sum^r_{j=1}u_jfe_{\a_j}$ $=\sum^t_{p=1}\left(\sum^r_{j=1}l_{pj}u_j\right)e_{\a_p}$.

Now, from the equation $y'=fy+g$, we obtain for $p=1,\cdots,r,$ that
\begin{equation}
u'_p+\sum^r_{j=1}n_{pj}u_j=\sum^r_{j=1}l_{pj}u_j+m_p.
\end{equation}
Consider the $k-$vector space $\mathcal{K}:=span_k\{1,u_1,\cdots,u_r\}$. From the above equation, it is clear that $\mathcal{K}$ is a (finite dimensional) differential $k-$vector space and that $k\subset \mathcal{K}\subset K$.  Then for every $u\in \mathcal{K}-k$, there exists a non negative integer $n$ such that  $u=u^{(0)}, u^{(1)},\cdots,u^{(n)}$  are linearly independent over  $k$ and that $u^{(n+1)}=\sum^n_{i=0}a_iu^{(i)}$, for some  $a_i\in k$.  Hence the lemma is proved.
\end{proof}

\begin{theorem}\label{solutiondense}
Let $E$ be a liouvillian extension field of $k$  and let $K$ be a differential field intermediate to $E$ and $k$.  Then \begin{itemize} \item [I.]$K=k\langle u_1,\cdots,u_l\rangle$ where for each $i$, $u_i$ satisfies a linear homogeneous differential equation over $k\langle u_1,\cdots, u_{i-1}\rangle$.\\ \item[II.] If $E=k(t_1,\cdots,t_n)$ is an iterated antiderivative extension of $k$ with $C_E=C_k$  then $K$ is an iterated antiderivative extension of $k$. That is, $K=k(v_1,\cdots,v_m)$, where $v'_i\in k(v_1,\cdots,v_{i-1})$ .\\   \item[III.] If $k$ is an algebraically closed field  then there is  an element $u\in K-k$ such that $u'=au+b$ for some $a,b\in k$.\\ \item[IV.]Finally, if $k$ is algebraically closed field and $c'=0$ for all $c\in k$,  then there is an element $z\in K-k$ such that either $z'=1$ or $z'=az$ for some element $a\in k$. \end{itemize}
\end{theorem}

\begin{proof}
We will use an induction on tr.d$(E|k)$ to prove item I.
For any differential field $k^*$ intermediate to $E$ and $k$ we observe that $E$ is a liouvillian extension of $k^*$ and therefore for differential fields $E\supset K\supset k^*$  such that tr.d$(E|k^*)<$tr.d$(E|k)$, we shall assume that the item I of our theorem holds.
If $u\in K$ is algebraic over $k$ then the vector space dimension $[k(u):k]=m$ for some positive integer $m$. Since $k(u)$ is a differential field,  the set $\{u, u',u'',\cdots, u^{(m)}\}$ is linearly dependent and therefore $u$ satisfies a linear differential equation over $k$. Since
 $k$ is of characteristic zero, there exist an element  $u_1\in K$ such that that $k(u_1)$ is the algebraic closure of $k$ in $K$.  Replacing $k$ with $k(u_1)$, if necessary, we shall assume that $k$ is algebraically closed in $K$ and that tr.d$(K|k)\geq 1$.
 Let $E=k(t_1,\cdots,t_n)$, $k_0:=k$ and choose the largest positive integer $m$ so that $k_{m-1}$ and $K$ are linearly disjoint over $k$. Since $k$ is  algebraically closed in $K$ and the characteristic of $k$ is zero, the field $k_{m-1}$ is algebraically closed in  $k_{m-1}(K)$. Now our choice of $m$ guarantees that $t_m$ is not algebraic over $k_{m-1}$ and that $t_m$ is algebraic over $k_{m-1}(K)$.   Let $\mathcal{P}(X)=\sum^n_{i=0}a_{n}X^n$ be the monic irreducible polynomial of $t_m$ over $k_{m-1}(K)$ and let $l$ be the smallest integer such that $a_{n-l}\notin k_{m-1}$. Expanding the equation $(\sum^n_{i=0}a_{i}t^i_m)'=0$  and comparing it with the equation $\sum^n_{i=0}a_{i}t^i_m=0$, we obtain

 $$a'_{n-l}= \begin{cases} -(n-(l-1))ra_{n-(l-1)} &\ \mbox{if }\  t'_m=r\in k_{m-1} \\
(n-l)ra_{n-l} &\ \mbox{if }\ t'_m/t_m=r\in k_{m-1}. \end{cases} $$

Thus there is an element $a_{n-l}\in k_{m-1}(K)-k_{m-1}$ such that  either $a'_{n-l} \in k_{m-1}$ or $a'_{n-l}/a_{n-l}\in k_{m-1}$. Now we apply lemma \ref{ldesol} and obtain an element $u_2\in K-k$ such that $L(u_2)=0$, where $L(Y)=0 $ is a linear  differential equation over $k$ of order $\geq 1$. Since $k$ is algebraically closed in $K$, such an element $u_2$ must be transcendental over $k$ and we now prove item I by setting $k^*=k\langle u_2\rangle$ and invoking the induction hypothesis.

Let $E$ be an iterated antiderivative extension of $k$ with $C_E=C_k$. We shall choose elements $t_1,\cdots,t_n\in E$ so that $E=k(t_1,\cdots,t_n)$ and that $t_1,\cdots,t_n$ are algebraically independent over $k$ (see \cite{Sri2010}, theorem 2.1).  Let $L(Y)=0$ be a linear homogeneous differential equation over $k$ of smallest positive degree $n$ such that $L(u)=0$ for some $u\in K-k$. Let $u\in k_{i}-k_{i-1}$ and observe that the ring $k_{i-1}[t_i]$ is a differential ring with no non trivial differential ideals (see \cite{Put-Sin}, example 1.18). Consider the differential $k-$vectorspace $W:=$span$_k\{u,u',\cdots,u^{(n-1)}\}$ and choose a nonzero  element $g\in k_{i-1}[t_i]$ so that $gu^{(j)}\in k_{i-1}[t_i]$  for all $j$. Then  the set $I=\{h\in k_{i-1}[t_i]\ |\ hW\subset k_{i-1}[t_i]\}$ is a non zero differential ideal of $k_{i-1}[t_i]$ and therefore $I=R$. Thus $1.u=u\in k_{i-1}[t_i]$ and let $u=a_pt^p_i+\cdots+a_0$, where $a_p\neq 0$ and for each $j$, $a_j\in k_{i-1}$. Then $L(a_pt^p_{i}+\cdots+a_0)=0$ together with the fact that $t_i$ is transcendental over $k_{i-1}$ implies $L(a_p)=0$ and thus we have found a non zero solution $a_p$ of $L(Y)=0$ in $k_{i-1}$. Repeating this argument, one can then show that there is a non zero element $d\in k$ such that $L(d)=0$. This means that $L(Y)=L_{n-1}(L_1(Y))$, where $L_1(Y)=Y'-(d'/d)Y$ and $L_{n-1}(Y)=0$ is a linear  homogeneous differential equation over $k$ of order $n-1$. Since $L_1(u)\in K$, from our choice of $u$ and $n$, we obtain that $L_1(u)\in k$. Let $v=u/d$ and observe that $v\in K-k$ and that $v'\in k$. Since $u\in k_i-k_{i-1}$, we know that $t_i$ is algebraic over $k_{i-1}(v)$ and since $C_E=C_k$, it follows from theorem \ref{construction} that  $t_i\in k_{i-1}(v)$. Thus $k(v)\subset K\subset E=k(v)(t_1,\cdots,t_{i-1},t_{i+1},\cdots,t_n)$ and an induction on the transcendence degree, as in the proof of I, will prove II.

Suppose that the differential field $k$ is an algebraically closed field. Let $L(Y)=0$ be a linear homogeneous differential equation over $k$ of smallest positive degree $n$ such that $L(u)=0$ for some $u\in K-k$. Then $L(Y)=0$ admits a liouvillian solution and  since $k$ is algebraically closed, the field of constants $C_k$ is algebraically closed as well. It follows from \cite{Sin1981}, theorem $2.4$ that  $L(Y)=L_{n-1}(L_1(Y))$, where  $L_{n-1}(Y)=0$ and $L_1(Y)=0$ are linear homogeneous differential equations over $k$ of degrees $n-1$ and $1$ respectively. Thus from the choice of $L$, we must have $L_1(u)\in k$.

Suppose that $k$ is an algebraically closed field with the trivial derivation. Assume that there is no $z\in K-k$ such that $z'=0$. We know from II that there is an element $u\in K-k$ such that $L(u)=0$, where $L(Y)=Y'-bY-c$ for some $b,c\in k$. If $b=0$ then let $z=(u/c)$ and observe that $z'=1$. Therefore we shall assume that $b\neq 0$. We will now show that there is an element $z\in k(u)-k$ such that $z'=mb z$ for some integer $m>0$.  Suppose  that there is no such integer $m$. Then we shall consider the field $E=k(u)(X)$, where $X$ is transcendental over $k(u)$ and define a derivation on $E$ with $X'=bX$. Then from  theorem \ref{construction} we obtain that  $C_E=C_{k(u)}$. But
$$\left(\frac{u}{X}\right)'=\left(\frac{-c}{bX}\right)'$$
and therefore $(bu+c)(1/bX)\in C_{k(u)}\subset k(u)$. Thus we obtain $X\in k(u)$, which is absurd. Hence the theorem is proved. \end{proof}

\begin{remark}\label{order2}

Let $E$ be an iterated antiderivative extension of $k$ with $C_E=C_k$. Then $E=k(t_1,\cdots,t_n)$, where $t_1,\cdots,t_n$ are algebraically independent over $k$ and furthermore, the differential field $E$ remains a purely transcendental extension over any of its differential subfields containing $k$ (see \cite{Sri2010}, theorems 2.1 \& 2.2). Suppose that  $y\in E-k$ and that $y'=\mathcal{P}(y)$ for some polynomial $\mathcal{P}$ in one variable over the differential field $k$. We will now show that deg $\mathcal{P}\leq 2$. The field $k(y)$ is a differential field intermediate to $E$ and $k$ and therefore, by theorem \ref{solutiondense}, there is an element $z\in k(y)-k$ such that $z'\in k$. Moreover, the element $y$ is algebraic over the differential subfield $k(z)$. Then since $E$ is a purely transcendental extension of $k(z)$, we obtain  that $y\in k(z)$. Thus $k(y)=k(z)$ and from  L{\"u}roth's theorem,  we know that there are elements $a,b,c,d\in k$ such that $ad-bc\neq 0$ and that $$z=\frac{ay+b}{cy+d}.$$
Let $\mathcal{P}(y)= a_ny^n+a_{n-1}y^{n-1}+\cdots+a_0$, $a_n\neq 0$ and $z'=r\in k$ and observe that \begin{equation*}(a'y+b'+a\mathcal{P}(y))(cy+d)-(c'y+d'+c\mathcal{P}(y))(ay+b)=r(cy+d)^2.\end{equation*}
Since $a_n\neq 0$ and $ad-bc\neq 0$, comparing the coefficients of $y^n$ in the above equation, we obtain that $n\leq 2$.  On the other hand, the differential equation $Y'=-rY^2$ has a  solution in $E-k$, namely, $1/z$ and thus  the bound $n\leq 2$ is sharp.

\end{remark}

\section{First Order  Non Linear Differential Equations}

Let $C$ be  an algebraically closed  field of characteristic zero and view $C$ as a differential field with the trivial derivation $c'=0$ for all $c\in C$.  Consider the field $C(X)$ of rational functions in one variable $X$. We are interested in the non constant liouvillian solutions of the following differential equations  \begin{enumerate} \item[(i)] $y'=\mathcal{R}(y)$, where $\mathcal{R}(X)\in C(X)$ is a non zero element and \item[(ii)] $(y')^2=\mathcal{P}(y),$ where $\mathcal{P}(X)\in C[X]$ and $\mathcal{P}(X)$ has no repeated roots in $C$.  \end{enumerate}

Suppose that there is a non constant element $y$ in some differential field extension  of  $C$ satisfying either (i) or (ii) then $C(y,y')$ must be a differential field, the element $y$ must be transcendental over $C$ and for any $z\in C(y)$, we have
\begin{equation}\label{two derivations} z'=y' \frac{\partial z}{\partial y}. \end{equation}

In the next proposition we will generalise a result of Singer (see \cite{Sin1975}, corollary 2) concerning elementary solutions of first order differential equations.
\begin{proposition}\label{M-B}
Let $C$ be  an algebraically closed  field of characteristic zero with the trivial derivation and let $\mathcal{R}(X)\in C(X)$ be a non zero element.  The equation $Y'=\mathcal{R}(Y)$  has a non constant solution $y$ which is liouvillian over $C$ if and only if there is an element $z\in C(y)$ such that
$$\dfrac{1}{\mathcal{R}(y)}\quad\text{is of the form}\quad \frac{\partial z}{\partial y} \ \ \text{or}\ \
										\dfrac{\frac{\partial z}{\partial y}}{a z}$$
for some non zero element $a\in C$.

\end{proposition}

\begin{proof}
 Let $y$ be a non constant liouvillian solution of $Y'=\mathcal{R} (Y)$. From equation \ref{two derivations}, we observe that if $z\in C(y)$ and $z'=0$ then $\partial z/\partial y=0$ and since the field of constants of $C(y)$ with the derivation $\partial/\partial y$ equals $C$, we obtain that $z\in C$. Since $y$ is liouvillian, the differential field $M=C(y)$ is contained in some liouvillian extension field of $C$ and therefore, applying  theorem \ref{solutiondense}, we  obtain an element $z\in C(y)-C$ such that either $z'=1$ or $z'=az$ for some (non zero) $a\in C$. Now it follows immediately from equation \ref{two derivations} that $1/\mathcal{R}(y)$ has the desired form.  Let $y'=\mathcal{R}(y)$ and $1/\mathcal{R}(y)$ equals $\partial z/\partial y$ or $(1/az)(\partial z/ \partial y)$ for some $z\in C(y)$ and for some non zero element $a\in C$. Then since $z'= y'(\partial z/\partial y)$,  we obtain $z'=1$ or $z'=az$.  From the fact that $C(y)\supset C(z)$, we see that $y$ is algebraic over the liouvillian extension $C(z)$ of $C$ and thus $C(y)$ is a liouvillan extension of $C$.  \end{proof}

\begin{proposition}\label{hyperellipticcurves}

Let $C$ be  an algebraically closed  field of characteristic zero with the trivial derivation. Let $\mathcal{P}(X)\in C[X]$ be a polynomial of degree $\geq 3$  with no repeated roots.  Then the differential equation $(Y')^2=\mathcal{P}(Y)$ has no non constant liouvillian solution over $C$. In particular, the elliptic function $y$ such that $(y')^2=y^3+ay+b$, where $\frac{a^3}{27}+\frac{b^2}{4}\neq 0$, is not  liouvillian.

\end{proposition}

\begin{proof}
Suppose that there is a non constant liouvillian solution $y$ satisfying   the equation $(Y')^2=\mathcal{P}(Y)$. Since $C$ is algebraically closed, such an element $y$ must be transcendental over $C$.  Applying theorem \ref{solutiondense} to the differential field $C(y,y')$, we obtain that there is an element $z\in C(y,y')-C$ such that either $z'=1$ or $z'=az$ for some  $a\in C$. We will first show that there is no $z\in C(y,y')-C$ with $z'=az$ for any $a\in C$.

For ease of notation, let $P=\mathcal{P}(y)$  and observe that $P$ is not a square in $C[y]$ and thus $y'\notin C(y)$ and therefore $y'$ lies in a quadratic extension of $C(y)$.
Write $z=A+By'$, where $A,B\in C[y]$ and $B\neq 0$. Then, taking derivatives, we obtain $A'+B'y'+By''=z'$ and using equation \ref{two derivations}, we  obtain  $$y' \frac{\partial A}{\partial y} + P\frac{\partial B}{\partial y}+  \frac{B}{2} \frac{\partial P}{\partial y}=z'.$$
If there is a non zero element $z\in C(y,y')$ such that $z'=az$ then, by comparing coefficients of the above equation, we obtain \begin{align}&\frac{\partial A}{\partial y}=aB\label{first eqn}\qquad\text{and}\\  &P\frac{\partial B}{\partial y}+  \frac{B}{2} \frac{\partial P}{\partial y}=aA\label{second eqn}.\end{align}
Multiplying the equation \ref{second eqn} by $2B$ and using \ref{first eqn}, we obtain $\partial (B^2 P)/\partial y=\partial A^2/\partial y. $
Thus there is a non zero constant $c\in C$ such that \begin{equation}\label{hypeleqn}B^2P=A^2+c.\end{equation} Write $A=A_1/A_2$ and $B=B_1/B_2$, where $A_1,A_2,B_1,B_2$ are polynomials in $C[y]$ such that  $A_1,A_2$ are relatively prime, $B_1,B_2$ are relatively prime  and $A_2,B_2$ are monic. Then since $P$ has no square factors, it follows from the equation $$B^2_1A^2_2P=A^2_1B^2_2+cA^2_2B^2_2$$ that $A_2=B_2$. The equation \ref{first eqn}, together with our assumption that $A_2$ and $B_2$ are monic, forces $A_2=B_2=1$ and we obtain deg $A $= 1+deg $B$. Now from equation \ref{hypeleqn}, we have $$2\ \text{deg}\ A -2+\ \text{deg}\  P=2\ \text{deg} \ A$$ and thus we obtain deg $ P=2$, which contradicts our assumption.

Now let us suppose that $z\in C(y,y')-C$ such that $z'=1$. Then  $\partial A/\partial y=0$, which imples $A\in C$ and we have $(\partial B/\partial y)P+  (B/2) (\partial P/\partial y)=1$. Thus  \begin{equation}\label{elliptic} \frac{\partial (B^2P)}{\partial y}=2B.\end{equation}
It is clear from the above equation that $B$ cannot be a  polynomial and that $B$ cannot have a pole of order $1$.   On the other hand if $c\in C$ is a pole of $B$ of order $m\geq 1$ then we shall write $B=R+\sum^m_{i=1}\b_i/(x-c)^i$, where $R\in C(y)$ and $c$ is not a pole of $R$. Since $P$ has no repeated roots, we conclude that $B^2P$ has  a pole at $c$ of order $\geq 2m-1$ and thus $c$ is a pole of $\partial (B^2P)/\partial y$  of order $\geq 2m$. This contradicts equation \ref{elliptic}.
\end{proof}

 Consider the ring of rational functions in two variables $C[Y,X]$. For constants $a\in C-\{0\}$ and $b\in C$, define a derivation $D$ of $C[Y,X]$ by setting $D(Y)=X$ and $D(X)=a/2$. Then $D(X^2-aY-b)=0$ and therefore the ideal $I=\langle X^2-(aY+b)\rangle$ is a differential ideal as well as a prime ideal. Thus the factor ring $C[Y,X]/I$ is a differential ring as well as a domain. Extend the derivation to the field of fractions $E$ of $C[Y,X]/I$.  Now, in the differential field $E$, we have  elements $x,y$ such that $y'=x$, $x'=a/2$ and $(y')^2=ay+b$. Let $z=2x/a$ and note that $z'=1$. Since tr.d$(E|C)=1$, the field $E$ is algebraic over $C(z)$ and thus $E$ is liouvillian over $C$. Thus the equation $(Y')^2=aY+b$ admits non constant liouvillian solutions. For the polynomial $f(X)=1-X^2\in \mathbb{C}(X)$, where $\mathbb{C}$ is the field of complex numbers, we have $(\frac{d}{dx}(\sin x))^2=f(\sin x)$ and thus $n\geq 3$ is necessary for the proposition \ref{hyperellipticcurves} to hold.

In the next proposition, we shall construct differential equations  whose liouvillian solutions, from liouvillian extensions having the same field of constants, are all algebraic over the ground field.

\begin{proposition}\label{algsolutions}
Let $\overline{k}$ be  an algebraic closure of $k$ and let $E$ be a  liouvillian extension of $\overline{k}$ with $C_E=C_{\overline{k}}$. Let $$F(Y,Y')=Y'-\a_nY^n-\cdots-\a_{2}Y^{n-1}-\a_1Y,$$
where $\a_1,\a_2,$ $\cdots,\a_n\in k$ and suppose that there is  an element $y\in E-\overline{k}$ such that  $F(y,y')=0$.
\begin{itemize}
\item [I.] If there is an element $\gamma\in \overline{k}$ such that $\gamma'=\a_1 \gamma$ then there is no $z\in \overline{k}(y)-\overline{k}$ such that $z'/z\in \overline{k}$.
\item[II.] If $\a_1=0$ and $x'\neq \a_2$ for all $x\in k$ then there is an element $w\in \overline{k}(y)-\overline{k}$ and an element $v\in \overline{k}$ such that $w'=\a_2$ and that $v'=\a_3$.

\end{itemize}

\end{proposition}

\begin{proof}
Suppose that there is an element $z\in\overline{k}(y)-\overline{k}$ such that $z'=\a z$ for some $\a\in \overline{k}$. Write $z=P/Q$ for some relatively prime polynomials $P=\sum^{s}_{i=0}a_iy^i$ and $Q=\sum^t_{j=0}b_jy^j$ in $\overline{k}[y]$ with $b_t=1$. Taking derivative, we obtain $\a PQ=P'Q-PQ'.$ Replacing $z$ and $\a$ by $1/z$ and $-\a$,  if necessary,  we shall assume that $b_0\neq 0$. Let $r$ be the smallest non zero integer such that $a_r\neq 0$.  Then we have  \begin{align}&\a(a_ry^{r}+ \cdots ) (b_0+\cdots)= (b'_0+\cdots)(a_ry^r+\cdots)\\&-((a'_r+ra_r\a_1)y^r+\cdots)(b_0+\cdots) \notag
\end{align}
 Now we compare the coefficients of $y^r$ and obtain that  $$b'_0a_r-a'_rb_0-\a a_rb_0=r\a_1 a_r b_0$$ and thus  $\left(\gamma^r a_r z/b_0\right)'=0.$ This contradicts our assumption that $C_{E}\neq C_{\overline{k}}$ and thus item I is proved.

Let us assume that $\a_1=0$ and that $x'\neq \a_2$ for all $x\in k$. From theorem \ref{solutiondense}, we obtain an element $z\in \overline{k}(y)-\overline{k}$ with $z'=\a z+\b$ for some $\a,\b\in \overline{k}$. We claim that there is an element $z_1\in \overline{k}(y)-\overline{k}$ with $z'_1\in \overline{k}.$ Write $z=P/Q$ for some relatively prime polynomials $P=\sum^{s}_{i=0}a_iy^i$ and $Q=\sum^t_{j=0}b_jy^j$ in $\overline{k}[y]$ with $b_t=1$. Taking derivative, we obtain \begin{equation}\label{Abel} \a PQ+\b Q^2=P'Q-PQ'.\end{equation}
Comparing the constant terms of the above equation, we have $\a a_0b_0+\b b^2_0=a'_0b_0-a_0b'_0$. If $b_0\neq 0$ then  $(a_0/b_0)'=\a(a_0/b_0)+\b$ and thus $[z-(a_0/b_0)]'=\a[z-(a_0/b_0)]$. This contradicts item I of the proposition.   Thus $b_0=0$ and  we choose the smallest integer $m$ such that $b_m\neq 0$ and observe that  $a_0\neq 0$. From equation \ref{Abel}, we have
\begin{align}\label{inhomo}&\a(b_my^{m}+ b_{m+1}y^{m+1}+\cdots ) (a_0+a_1y+\cdots)+\b(b^2_my^{2m}+ \\ & 2b_mb_{m+1}y^{2m+1}+\cdots ) = (a'_0+a'_1y+\cdots)(b_my^m+\cdots)
-(b'_my^m +\notag\\ &(mb_m\a_2+b'_{m+1})y^{m+1}+\cdots)(a_0+a_1y+\cdots) \notag
\end{align}
Compare the coefficients of $y^m$ and obtain $\a a_0b_m=a'_0b_m-a_0b'_m$.   Therefore $(a_0/b_m)'=\a (a_0/b_m)$ and since   $z'=\a z+\b$, it follows that
\begin{equation} \label{antipresent}\left(\frac{b_mz}{a_0}\right)'= \frac{\b b_m}{a_0}.\end{equation}
Taking $z_1=b_mz/a_0$, we prove our claim.  Thus, in the above calculations,  we shall suppose that $\a=0$. Then we have \begin{align}\label{antieqn}& \b(b^2_my^{2m}+ 2b_mb_{m+1}y^{2m+1}+\cdots )=(a'_0+ a'_1y+\cdots)(b_my^m+  b_{m+1}y^{m+1}\\ &+\cdots)-  (a_0+a_1 y+\cdots) (b'_my^m+(mb_m\a_2+b'_{m+1})y^{m+1}+\cdots) \notag
\end{align}
Equating the coefficients of $y^m$, we obtain  $a'_0b_m-b'_ma_0=0$ and thus  for some non zero  $c\in C_{\overline{k}}$, we have $ca_0=b_m$. Note that if $m\geq 2$ then comparing the  coefficient  of $y^{m+1}$, we obtain \begin{equation}\label{inhomo2}a'_0b_{m+1}-a_0b'_{m+1}+a'_1b_m-a_1b'_m=mb_m\a_2a_0\end{equation} and substituting $b_m=ca_0$, we obtain that $[(b_{m+1}+ca_1)/(mca_0)]'=\a_2$. Since $(b_{m+1}+ca_1)/(mca_0)\in \overline{k}$, we shall apply theorem \ref{construction} and obtain a  contradiction to our assumption on $\a_2$. Therefore $m=1$ and from equation \ref{antieqn}, we have $f'=\a_2+c\b$, where $f=(b_{2}+ca_1)/(ca_0)$. Then $(f-cz)'=\a_2$, where $f-cz\in \overline{k}(y)$ and $f-cz\notin \overline{k}$.  This proves the first part of  item II.

Now, in all the above calculations, we shall assume that $\a=0$ and $\b=\a_2$. Then $f'=(1+c)\a_2$ and note that if $c\neq -1$ then $(f/(1+c))'=\a_2$ and again we obtain a contradiction to our assumption on $\a_2$. Thus $c=-1$.  Now, $f'=0$ and consequently we have $a_1-b_2=c_1a_0$ for some constant $c_1\in C_{\overline{k}}$. We compare the coefficients of $y^3$ and obtain \begin{align}a'_2b_1-a_2b'_1+a'_1b_2-a_1b'_2+a'_0b_3-a_0b'_3= 2b_1b_2\a_2+2a_0b_2\a_2+a_0b_1\a_3\notag.\end{align} Substituting $b_1=-a_0$ and $b_2=a_1-c_1a_0$ in the above equation, we obtain
\begin{align}-a'_2a_0+a_2a'_0-c_1a'_1a_0+c_1a_1a'_0+a'_0b_3-a_0b'_3= -a^2_0\a_3\notag.\end{align}
Then it is easy to see that  $v'=\a_3$ for $v=(a_2+c_1a_1+b_3)/a_0\in \overline{k}$.  \end{proof}

\subsection*{Final Comments}  One can slightly extend the proposition \ref{algsolutions} to include differential equations of the form  $y'=\a_ny^n+\cdots+\a_3y^3+\a_2y^2+\a_1y$, when $\gamma'/\gamma=\a_1$ for some non zero $\gamma\in k$ and  $x'\neq \gamma \a_2$ and $x'\neq \gamma^{2}\a_3$ for any $x\in k$. The proof that there are no elements in $E-\overline{k}$ satisfying the differential equation follows immediately from proposition \ref{algsolutions} once we observe that $(y/\gamma)'=$ $\gamma^{n-1}\a_n(y/\gamma)^n+\cdots+\gamma^2\a_3(y/\gamma)^3+\gamma\a_2(y/\gamma)^2$. Let $\mathbb{C}(x)$ be the field of rational functions in one variable over the field of complex numbers with the usual derivation $'=d/dx$. It is evident that $Y=-x$ is a liouvillian solution of
the  differential equation \begin{equation}\label{abeleqn}Y'= \frac{1}{x^3} Y^3+\frac{1}{x^2} Y^2+\frac{1}{x} Y.\end{equation}
Since $x'/x=1/x$, we see that $y$ is a solution of equation \ref{abeleqn} if and only if $y/x$ is a solution of $Y'=(1/x)(Y^3+Y^2)$. Now it follows from proposition \ref{algsolutions} that equation \ref{abeleqn} has no solution in $E-\overline{\mathbb{C}(x)}$ for any  liouvillian extension  $E$ of  $\overline{\mathbb{C}(x)}$ with $C_E=C_{\overline{\mathbb{C}(x)}}$.

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