\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{graphicx}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 66, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/66\hfil Special solutions of the Riccati equation]
{Special solutions of the Riccati equation with applications
to the Gross-Pitaevskii nonlinear PDE}

\author[A. Al Bastami, M. R. Beli\'c, N. Z. Petrovi\'c \hfil EJDE-2010/66\hfilneg]
{Anas Al Bastami, Milivoj R. Beli\'c, Nikola Z. Petrovi\'c}  % in alphabetical order

\address{
Science Program, Texas A\&M University at Qatar, P.O. Box 23874
Doha, Qatar}
\email[A. Al Bastami]{anas.al\_bastami@qatar.tamu.edu}
\email[M. R. Beli\'c]{milivoj.belic@qatar.tamu.edu}
\email[N. Z. Petrovi\'c]{nikola.petrovic@qatar.tamu.edu}

\thanks{Submitted April 6, 2010. Published May 8, 2010.}
\subjclass[2000]{34A34, 34A05}
\keywords{Riccati equation; solutions to Gross-Pitaevskii equation;
\hfill\break\indent chirp function}

\begin{abstract}
 A method for finding solutions of the Riccati differential equation
 $y' = P(x) + Q(x)y + R(x)y^2$ is introduced. Provided that certain
 relations exist between the coefficient $P(x)$, $Q(x)$ and
 $R(x)$, the above equation can be solved in closed form.
 We determine the required relations and find the general solutions
 to the aforementioned equation. The method is then applied to the
 Riccati equation arising in the solution of the multidimensional
 Gross-Pitaevskii equation of Bose-Einstein condensates by the
 F-expansion and the balance principle techniques.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}

\section{Introduction}

The Riccati equation (RE), named after the Italian mathematician
Jacopo Fran\-cesco Riccati \cite{wiki}, is a basic first-order
nonlinear ordinary differential equation (ODE) that arises in
different fields of mathematics and physics \cite{ricc}. It has the
form
\begin{equation} \label{e1}
y' = P(x) + Q(x)y + R(x)y^2,
\end{equation}
which can be considered as the lowest order \emph{nonlinear}
approximation to the derivative of a function in terms of the
function itself. It is
assumed that $y$, $P$, $Q$ and $R$ are real functions of the real
argument $x$. It is well known that solutions to the general Riccati equation are
not available, and only special cases can be treated
\cite{ince,davis,reid,kamke,zwil,pol}. Even though the equation is
nonlinear, similar to the second order inhomogeneous linear ODEs one
needs only a particular solution to find the general solution.

In a standard manner Riccati equation can be reduced to a second-order
linear ODE
\cite{wiki,ince} or to a Schr\"odinger equation (SE) of quantum
mechanics \cite{schwabl}. In fact, Riccati equation naturally arises in many
fields of quantum mechanics; in particular, in quantum chemistry
\cite{fraga}, the Wentzel-Kramers-Brillouin approximation \cite{wkb}
and SUSY theories \cite{susy}. Recently, methods for solving the
Gross-Pitaevskii equation (GPE) arising in Bose-Einstein condensates
(BECs) \cite{atre,yang} based on Riccati equation were introduced. Our objective
is to find new solutions of Riccati equation by utilizing relations between the
coefficient functions $P(x)$, $Q(x)$ and $R(x)$ for which the above
equation can be solved in closed form.

It is well known that any equation of the Riccati type can always be
reduced to the  second order linear ODE
\begin{equation} \label{e2}
u'' - \big[ Q(x) + \frac{R'(x)}{R(x)} \big] u' + P(x)R(x)u =
0
\end{equation}
by a substitution $y=-u'/(uR)$. It is also known that if one can
find a particular solution $y_p$ to the original equation, then the
general solution can be written as $y = y_p + {1}/{w}$
\cite{mathworld1}, where $w$ is the general solution of an
associated linear ODE
\begin{equation} \label{e3}
w' + \left[ Q(x) + 2R(x)y_p \right] w + R(x) =
0\end{equation}
which does not contain $P(x)$. Solving this equation
we get \cite{eqworld}: $w = w_0e^{-\phi (x)} - e^{-\phi
(x)}\int_{x_0}^x R(\xi) e^{\phi (\xi)} \,d\xi$, where
$\phi (x) = \int_{x_0}^x [Q(\xi) + 2R(\xi)y_p] \,d\xi$.
It is clearly seen from the relation above that $w_0 = \frac{1}{y_0
- y_{p0}}$. The general solution is therefore given by
\cite{eqworld}:
\begin{equation} \label{e4}
y = y_p + e^{\phi(x)}\big[\frac{1}{y_0 - y_{p0}} -
\int_{x_0}^x R(\xi) e^{\phi (\xi)} \,d\xi\big]^{-1}.
\end{equation}

This article contains four sections. Section 2 introduces the solution
method, Sec. 3 presents an application and Sec. 4 brings a
conclusion.

\section{Solution method}

Equation \eqref{e1} cannot be solved in closed form for arbitrary
functions $P(x)$, $Q(x)$ and $R(x)$. However, if certain relations
exist between these functions, then the above equation can be
transformed into a second order linear ODE, which can be easily
solved in two cases: If it contains constant coefficients, or if it
contains certain coefficient functions.

For the sake of making our calculations clearer, we make the
following two substitutions: $a(x) = -( Q + R'/{R})$ and $b(x) =
P(x)R(x)$. Now the above ODE for $u$
becomes\begin{equation} \label{ux}
\frac{d^2u}{dx^2} +
a(x)\frac{du}{dx} + b(x)u = 0.\end{equation} Consider an arbitrary
function of $x$, $z \equiv f(x)$, which we choose to be a new
independent variable. The substitution looks arbitrary, but it will
be made more specific in a moment. We compute the first and second
derivatives of $u$ with respect to $x$, but now in terms of the new
independent variable $z$:
\begin{gather}
\frac{du}{dx} = \frac{du}{dz}\frac{dz}{dx} \label{e6}\\
\frac{d^2u}{dx^2} = \frac{d^2z}{dx^2}\frac{du}{dz} +
\big(\frac{dz}{dx}\big)^2\frac{d^2u}{dz^2}. \label{e7}
\end{gather}
We plug the last results into the differential equation (\ref{ux}),
to get:
\begin{equation} \label{e8}
\Big(\frac{dz}{dx}\Big)^2\frac{d^2u}{dz^2} +
\Big[\frac{d^2z}{dx^2} + a(x)\frac{dz}{dx}\Big]\frac{du}{dz} +
b(x)u = 0.
\end{equation}
Finally, dividing by $\big(\frac{dz}{dx}\big)^2$, we obtain
\cite{mathworld2}:
\begin{gather}
\frac{d^2u}{dz^2} + \Big[\frac{\frac{d^2z}{dx^2} +
a(x)\frac{dz}{dx}}{\big(\frac{dz}{dx}\big)^2}\Big]\frac{du}{dz} +
\Big[\frac{b(x)}{\big(\frac{dz}{dx}\big)^2}\Big]u = 0 \label{e9}\\
        \equiv \frac{d^2u}{dz^2} + 2{A}\frac{du}{dz} + B u =0, \label{e10}
\end{gather}
provided ${dz}/{dx}$ is not equal to 0.
The obtained equation can easily be solved in closed form if $A$ and
$B$ are either constants \cite{mathworld2} or if they are some
special functions for which the closed-form solutions to \eqref{e10}
are known. In this paper we consider only the two special cases,
namely when $A$ and $B>0$ are constants, or when $A = 0$ and $B$ is
an arbitrary function $B(x)$.

If $b(x)$ is positive, by considering the coefficient of $u$ we
define $z$ to be the following function:
\begin{equation} \label{e11}
z \equiv z_0 + s\int_{x_0}^x \sqrt{\frac{b(\xi)}{B}} d\xi,
\end{equation}
where $s=\pm 1$. The requirement that $b(x)$ is positive is
equivalent to the condition that the product $P(x)R(x)$ is positive.
To simplify bookkeeping, let $c=b/B$; then we have the following
relations:
\begin{gather}
\frac{dz}{dx} = sc^{1/2}, \label{e12} \\
\frac{d^2z}{dx^2} = \frac{c'}{2sc^{-1/2}}. \label{e13}
\end{gather}
 From \eqref{e12} it is clear that ${dz}/{dx}$ cannot be equal to
0. Now we compare the coefficients of ${du}/{dz}$ and use relations
\eqref{e12} and \eqref{e13} to get:
\begin{equation} \label{e14}
\frac{c'}{2sc^{1/2}} + asc^{1/2} - 2Ac = 0,
\end{equation}
or
\begin{equation} \label{e15}
\big(\frac{b}{B}\big)' + 2a\big(\frac{b}{B}\big) -
4As\big(\frac{b}{B}\big)^{3/2}=0.
\end{equation}
At this point it is more convenient to consider the two
cases separately.

\subsection{Case 1: $A$ and $B$ are constants}

If \eqref{e10} has constant coefficients $2A$ and $B$, then it is
easily solvable in closed form. This means:
\begin{equation} \label{e16}
b' + 2ab - \frac{4sA}{\sqrt{B}}b^{3/2}=0\end{equation}
or:
\begin{equation} \label{e17}
\frac{b'(x) + 2a(x)b(x)}{[b(x)]^{3/2}} = \frac{4sA}{\sqrt{B}}.
\end{equation}
Substituting back the original
expressions for $a(x)$ and $b(x)$, we get the final result:
\begin{equation} \label{e18}
\frac{\left[ P(x)R(x) \right]' - 2\left[ Q(x) + {R'(x)}/{R(x)}
\right]P(x)R(x)}{[P(x)R(x)]^{3/2}} = \frac{4sA}{\sqrt{B}}.
\end{equation}
At this point a few comments are in order. First, note that we are
stating that if the condition \eqref{e18} is satisfied, then the
general solution can be found. However, when the condition is
\emph{not} satisfied, this does not mean that the general solution
cannot be found. In fact, most of the known special cases of Riccati
equation (with known solutions) \cite{pol} do not satisfy the
relation obtained.

Second, one may object that in place of the original nonlinear Riccati equation we
obtained another \emph{nonlinear} equation for $b(x)$, which might be
equally difficult to solve. Luckily, this is not the case; \eqref{e16}
has a constant coefficient in front of the nonlinear term (which is
also a \emph{variable} parameter at our will) and hence is more
manageable. It often allows easy solutions, as we display below, for
which one can find nontrivial solutions of the original Riccati equation.

Now we proceed to solve \eqref{e10}. The general solution is given by:
\begin{equation} \label{e19}
u(x) = c_1e^{\lambda_1z} + c_2e^{\lambda_2z},
\end{equation}
where $z$ is the function defined in \eqref{e11}, $c_1$ and $c_2$ are
some initial values, and $\lambda_1$ and $\lambda_2$ are the roots
of the characteristic polynomial $\lambda^2 + 2A\lambda + B=0$,
given by:
\begin{equation} \label{e20}
\lambda_{1,2} = -{A} \pm \sqrt{{A}^2 - B}.
\end{equation}
Hence, we assume that ${A}^2 \geq
B>0$, so that both lambdas are real and negative. This condition is
not necessary for the solution procedure, but is convenient for the
applications of solutions, which require real functions. We need
only a particular solution of \eqref{e10}, so we consider only
$u_p = e^{\lambda z}$, where $\lambda$ is any of the roots to the
polynomial.

 From the substitution done in \eqref{e2}, namely $y = -{u'}/[{uR(x)}]$,
we find the particular solution to be:
\begin{equation} \label{e21}
y_p = -\frac{s\lambda}{\sqrt{B}}\sqrt{\frac{P(x)}{R(x)}}
\end{equation}
Finally, we plug $y_p$ into the expression for the general solution
of Riccati equation, to find:
\begin{equation} \label{e22}
y = -\frac{s\lambda}{\sqrt{B}}\sqrt{\frac{P(x)}{R(x)}} +
e^{\phi(x)}\Big[\frac{1}{y_0 +
\frac{s\lambda}{\sqrt{B}}\sqrt{\frac{P(0)}{R(0)}}} - \int_{x_0}^x
R(\xi) e^{\phi (\xi)} \,d\xi\Big]^{-1}
\end{equation}
Note that we have substituted $y_{p0}$ by its value. To
recapitulate, here $A$ and $B$ are two arbitrary constants
satisfying $A^2 \geq B>0$, $\lambda$ is one of the roots of the
characteristic polynomial, $y_0$ is the initial condition for $y$,
and
\begin{equation} \label{e23}
\phi (x) = \int_{x_0}^x \Big[Q(\xi) -
\frac{2s\lambda}{\sqrt{B}}\sqrt{P(\xi)R(\xi)}\Big] \,d\xi
\end{equation}
is the integrating exponent. Below we apply this
general result to some specific examples.

\subsection{Case 2: $A=0$ and $B=B(x)$}

When $A=0$, \eqref{e14} reduces to the simple equation $c' = -2ac$.
Solving for $c$, and remembering that $c=b/B$, we get the
simple relation
\begin{equation} \label{e24}
\frac{b}{B} = \Big(\frac{b}{B}\Big)_0
\exp\Big(-2\int_{x_0}^x a \,dx\Big),
\end{equation}
where $a(x)$ and $b(x)$ are given by the
original Riccati equation, and $B(x)$ is still an arbitrary function. Note that
\eqref{e10} now becomes
\begin{equation} \label{e25}
\frac{d^2u}{dz^2} + B(z) u =0,
\end{equation}
where $z$ is given by \eqref{e11}. When $B(z)$ is chosen as
$B(z)=B_0+B_1(z)$, then the last equation becomes equivalent to the
Schr\"odinger equation of quantum mechanics, which is a linear second order differential
equation of the form:
\begin{equation} \label{e26}
\psi'' + \frac{2m}{\hbar^2}(E-V)\psi=0.
\end{equation}
This is an equation for
the wave function $\psi=u(z)$ of a particle of mass $m$ moving in a
potential $V=-{\hbar^2}B_1(z)/2m$ with an energy eigenvalue
$E={\hbar^2}B_0/2m$ \cite{liboff}. There are many specific
potentials $V$ for which the solutions $\psi_n$ and the energies
$E_n$ in the above equation are known. Here $n$ denotes some set of
quantum numbers. Therefore, one can choose $B(z)$ such that the
solutions $u_n(z)$ can be found. If $u_n(z)$ (and hence $u_n(x)$)
are known, then the solutions $y_n$ to Riccati equation can be easily written down
from the substitution $y_n(x)=-u_n'/(u_nR)$ mentioned above. This in
fact gives rise to various solutions of the various special cases of
Riccati equation.


\section{Application}

In \cite{petr} we considered the generalized GPE in (3+1)D for the
BEC wave function $u(x,y,z,t)$, with distributed time-dependent
coefficients \cite{atre,yang,book1}:
\begin{equation} \label{gen}
i\partial_t u + \frac{\beta(t)}{2}\Delta u + \chi(t)|u|^{2} u+
\alpha(t) r^2 u = i\gamma(t)u.
\end{equation}
Here $t$ is time, $\Delta
=\partial_x^{2}+\partial_y^{2}+\partial_z^{2}$ is the 3D Laplacian,
$r=\sqrt{x^2+y^2+z^2}$ is the position coordinate, and $\alpha(t)$
stands for the strength of the quadratic potential as a function of
time. The functions $\beta$, $\chi$ and $\gamma$ stand for the
diffraction, nonlinearity and gain/loss coefficients, respectively.

According to the F-expansion and the balance principle techniques
\cite{ref11}, in \cite{petr} we sought the solution in the form:
\begin{equation} \label{e28}
u(x,y,z,t) = \mathcal{M}(x,y,z,t) \exp{[i\mathcal{P}(x,y,z,t)]},
\end{equation}
where the magnitude $\mathcal{M}(x,y,z,t)$ and the phase
$\mathcal{P}(x,y,z,t)$ are given by
\begin{gather}\label{sol}
\mathcal{M}(x,y,z,t) = f(t)F(\theta)+g(t)F^{-1}(\theta),\\
\theta = k(t)x + l(t) y + m(t) z+\omega(t),  \label{e30}\\
\label{solend}
\mathcal{P}(x,y,z,t) = a(t)r^{2}+ b(t) (x + y + z) + e(t).
\end{gather}
Here $f$, $g$, $k$, $l$, $m$, $\omega$, $a$, $b$, $e$ are parameter
functions to be determined, and $F$ is one of the Jacobi elliptic
functions. The functions $a(t)$ and $b(t)$ should not be confused
with the functions $a(x)$ and $b(x)$ used before. Of all the
parameters, by far the most important is the chirp function $a(t)$,
because all other parameters, as well as the general solution of
GPE, can be expressed in terms of $a$. On the other hand, the
equation for the determination of $a$ is a Riccati equation of
the following type
\cite{petr}:
\begin{equation} \label{e32}
\frac{da}{dt} + 2\beta (t) a^2 - \alpha (t)=0.
\end{equation}
To this equation we apply the method developed in
this paper. We take $A$ and $B$ to be constant here.

Put in the form of the original Riccati equation, the coefficients are:
\begin{equation} \label{e33}
P(t) = \alpha(t),\quad Q(t) = 0,\quad R(t) = -2\beta(t).
\end{equation}
We write down relation \eqref{e18} between $\alpha$ and $\beta$ for
which \eqref{e32} is solvable in closed form:
\begin{equation} \label{e34}
\frac{\alpha \beta' -\alpha'\beta}{(-\alpha \beta)^{3/2}} =
\frac{4\sqrt{2}sA}{\sqrt{B}}.
\end{equation}
The prime is now the derivative with respect to $t$. Equation
\eqref{e34} can be manipulated to become a simple differential
equation for $-\alpha/\beta$:
\begin{equation} \label{e35}
\frac{\big(-\frac{\alpha}{\beta}\big)'}
{\big(-\frac{\alpha}{\beta}\big)^{3/2}\beta}
= \frac{4\sqrt{2}sA}{\sqrt{B}}.
\end{equation}
Solving this equation, one finds:
\begin{equation} \label{e36}
\sqrt{-\frac{\beta}{\alpha}} =
\sqrt{-\frac{\beta_0}{\alpha_0}} - \frac{2\sqrt{2}sA}{\sqrt{B}}
\int_0^t \beta \,dt.
\end{equation}
 Now one can write
down the solution for $a(t)$ from \eqref{e22}, provided the above
condition is satisfied:
\begin{equation} \label{e37}
a(t) = -\frac{s\lambda}{\sqrt{B}}\sqrt{-\frac{\alpha (t)}{2\beta
(t)}} + e^{\phi(t)}\Big[\frac{1}{a_0 +
\frac{s\lambda}{\sqrt{B}}\sqrt{-\frac{\alpha_0}{2\beta_0}}} +2
\int_0^t \beta (\tau) e^{\phi (\tau)} \,d\tau\Big]^{-1},
\end{equation}
where
$\phi (t) = -{2\sqrt{2}s\lambda}\int_0^t
\sqrt{-\alpha (\tau)\beta (\tau)} \,d\tau/{\sqrt{B}}$.
Note that the $-$ sign in the
square root indicates that $\alpha$ and $\beta$ have to be of the
opposite signs, which is consistent with the requirement that the
original function $b$ is positive. Hence, as long as the ratio of
the diffraction coefficient to the strength of the parabolic
potential can be made to satisfy \eqref{e35}, one can write down the
exact solutions to GPE. It should be mentioned that these functions
are the material parameters in BECs that are accessible to
experimental manipulation.

Our solution method for the GPE requires that $\beta$ be
proportional to $\chi$, and $\chi$ in turn be proportional to the
s-wave scattering length \cite{yuce}. To validate our proposed
solution method, we present a couple of examples in which $\beta$,
and hence the scattering length, are given by some representative
functions of time. In all the examples we determine the
corresponding chirp functions $a(t)$, from which one can write down
the exact solutions of the GPEs in question \cite{petr}. To avoid
singularities that are likely to appear in $\alpha(t)$ and $a(t)$ we
are choosing $s$ to be $-1$. Note that the appearance of
singularities is not detrimental to our method or to the theory of
BECs based on GPE, because that model is known to be valid only on a
limited time interval.

\subsection{Example 1: $\beta = \frac{1}{2}(e^{-\delta t}+1)$}
We consider first the case when $\beta$ is an exponential function
of time, $\beta(t) = \frac{1}{2}(e^{-\delta t}+1)$, where $\delta$
is some arbitrary parameter. This function describes a smooth change
in $\beta(t)$ from 1 to 1/2. First, \eqref{e36} is solved for $\alpha$,
to obtain:
\begin{equation} \label{e38}
\alpha(t) = -\frac{1+e^{-\delta t}}{2 \big(1+\frac{\sqrt{2}
(1-e^{-\delta t}+ \delta t)}{\delta }\big)^2}
\end{equation}
Then one finds $\phi$:
\begin{equation} \label{e39}
\phi(t) =  \delta t +\ln\big|\frac{\delta }{-\sqrt{2}+e^{\delta t}
\left(\sqrt{2}+\delta +\sqrt{2} \delta
t\right)}\big|.
\end{equation}
 Taking $\alpha_0 = -1$, $A=B=1$, and
performing the calculations, we obtain the following solution for
$a$:
\begin{equation} \label{e40}
a(t) = \frac{-\delta  }{2-2 e^{- \delta t}+\sqrt{2} \delta
+2 t \delta }-\frac{\delta  \sqrt{2} e^{ \delta t}}
{\left[-\sqrt{2}+e^{ \delta t} \left(\sqrt{2}+\delta
+\sqrt{2} t \delta \right)\right] \zeta(t)}
\end{equation}
where
$$
\zeta(t) = \delta t +\ln\big|\frac{\delta }{-\sqrt{2}+e^{t
\delta } (\sqrt{2}+\delta +\sqrt{2} t \delta
)}\big|-\frac{2}{1+\sqrt{2} a_0}.
$$
Although this solution
looks complicated, it allows simple expressions in the limit
$\delta\rightarrow 0$, when $\beta$ becomes constant. Figure \ref{fig1}
presents some representative cases of $\alpha$ and $a$ functions for
different values of $\delta$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig1}
\end{center}
\caption{Graphs of (a): $\alpha(t)$, (b): $a(t)$ for $a_0 = 0$, and
(c): $a(t)$ for $a_0 = 1$, for $\delta = 0.01,0.1,1,10$ (top to
bottom).}
\label{fig1}
\end{figure}

\subsection{Example 2: $\beta = \sum_{n=0}^N \beta_n t^n$}
Next, we consider the case when $\beta$ is some power series of the
form $\sum_{n=0}^N \beta_n t^n$, where $\beta_0\ne0$.
We go through the same procedure and solve \eqref{e36} for $\alpha$,
to get:
\begin{equation} \label{e41}
\alpha(t) = -\frac{\sum_{n=0}^N \beta_n t^n}{
\Big(1+2\sqrt{2}\sum_{n=0}^N \beta_n \frac{t^{n+1}}{n+1}\Big)^2}.
\end{equation}
Then we find $\phi$ to be:
\begin{equation} \label{e42}
\phi(t)=\ln \frac{1}{\big|1+2\sqrt{2}\sum_{n=0}^N
\beta_n \frac{t^{n+1}}{n+1}\big|}.
\end{equation}
 Again, taking
$\alpha_0 = -1$, $A=B=1$, and performing the calculations, we arrive
at the following closed-form solution:
\begin{equation} \label{e43}
a(t) = \frac{2\sqrt{2}a_0-(a_0\sqrt{2}+1)\ln \big|1+2\sqrt{2}
\sum_{n=0}^N \beta_n \frac{t^{n+1}}{n+1}\big|}
{\big(1+2\sqrt{2}\sum_{n=0}^N
\beta_n \frac{t^{n+1}}{n+1}\big)\big[2\sqrt{2}+(2a_0+\sqrt{2})\ln
\left|1+2\sqrt{2}\sum_{n=0}^N \beta_n
\frac{t^{n+1}}{n+1}\right|\big]}.
\end{equation}
These solutions
for $\alpha$ and $a$ are plotted in Fig. \ref{fig2}. Note that by choosing
different parameters $\beta_n$ and letting $N\rightarrow\infty$ one
can obtain closed-form expressions for different functions
$\beta(t)$. Figure \ref{fig3} presents the case with $\beta=\cos(\Omega t)$.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig2}
\end{center}
\caption{Same as Fig. \ref{fig1}. (a) $\alpha(t)$, (b) $a(t)$ for $a_0 = 0$,
and (c) $a(t)$ for $a_0 = 1$. Parameters: $N = 0,1,2,3,4$ (top to
bottom at $t=0.5$ for $\alpha$; bottom to top at $t=3$ for $a$),
$\beta_n=1$.}
\label{fig2}
\end{figure}

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.95\textwidth]{fig3}
\end{center}
\caption{Same as Fig. \ref{fig2}, but for $\beta(t)=\cos(\Omega t)$. (a)
$\alpha(t)$, (b) $a(t)$ for $a_0 = 0$, and (c) $a(t)$ for $a_0 = 1$.
Here $\Omega=6,7,8,9,10$; Curves with higher peaks correspond to
lower values of $\Omega$.}
\label{fig3}
\end{figure}


\subsection{Example 3: $\beta =
\tilde\beta\big(1-\frac{D}{B_1t-B_0}\big)$}
Finally, we consider the case when $\beta$ is of the form shown
above. This form is dictated by the dependance of the scattering
length on the magnetic field near the Feshbach resonance of cold BEC
atoms \cite{yuce}. The magnetic field $B(t)=B_1t$ (again, not to be
confused with the function $B(x)$ from the solution procedure) is
assumed to be linearly ramped in time near the resonance field
$B_0$. The parameter $D$ stands for the width of the resonance. Such
a dependence is found relevant not only on theoretical grounds
\cite{yuce} but most importantly experimentally \cite{inou}.

The closed-form solution is again readily obtained; however, this
time it includes integrals that cannot be evaluated in terms of
elementary functions. The results for $\alpha$, $\phi$, and $a$ are
as follows:
\begin{gather} \label{e44}
\alpha(t) = -\tilde\beta \frac{1-\frac{D}{B_1t-B_0}}
{\big[1-2\sqrt{2}\tilde\beta t +
\frac{2\sqrt{2}\tilde\beta
D}{B_1}\ln\big|\frac{B_1t-B_0}{B_0}\big|\big]^2},
\\ \label{e45}
\phi(t) = 2\sqrt{2}\hspace{1 mm}\tilde\beta\int_0^t
 \frac{1-\frac{D}{B_1\tau-B_0}}{1-2\sqrt{2}\tilde\beta \tau +
\frac{2\sqrt{2}\tilde\beta
D}{B_1}\ln\big|\frac{B_1\tau-B_0}{B_0}\big|}\,d\tau,
\\ \label{e46}
a(t) = \frac{1}{\sqrt{2}-2\tilde\beta t
+ \frac{4\tilde\beta D}{B_1}\ln\big|\frac{B_1t-B_0}{B_0}\big|} +
\frac{e^{\phi(t)}}{\frac{\sqrt{2}}{a_0\sqrt{2} -1} +
2\tilde\beta\int_0^t
\big(1-\frac{D}{B_1\tau-B_0}\big)e^{\phi(\tau)} \,d\tau}.
\end{gather}

\section{Conclusion}

We conclude the paper by restating our results. Provided for the
case 1 that the following condition between the coefficient
functions of Riccati equation $P(x)$, $Q(x)$, and $R(x)$ is met:
\begin{equation} \label{e47}
\frac{\left[ P(x)R(x) \right]' - 2\left[ Q(x) +
{R'(x)}/{R(x)} \right]P(x)R(x)}{[P(x)R(x)]^{3/2}} =
\frac{4sA}{\sqrt{B}},
\end{equation}
then the general solution of Riccati equation is given by
 \begin{equation} \label{e48}
 y = -\frac{s\lambda}{\sqrt{B}}\sqrt{\frac{P(x)}{R(x)}} +
e^{\phi(x)}\Big[ \frac{1}{y_0 +
\frac{s\lambda}{\sqrt{B}}\sqrt{\frac{P(0)}{R(0)}}} - \int_{x_0}^x
R(\xi) e^{\phi (\xi)} \,d\xi\Big]^{-1}.
\end{equation}
Here $A$ and $B$ are two arbitrary constants satisfying
${A}^2 \geq B>0$; $\lambda = -{A} \pm \sqrt{{A}^2 - B}$ is one
of the two roots
of the characteristic polynomial; $y_0$, $P(0)$, $R(0)$ are the
given boundary conditions; and
$\phi (x) = \int_{x_0}^x \big[Q(\xi)
- \frac{2s\lambda}{\sqrt{B}}\sqrt{P(\xi)R(\xi)}\big] \,d\xi$.


In the other case when $A=0$ and $B$ is some arbitrary function of
$x$, provided the following relation between the coefficient
functions $a$ and $b$ is valid:
\begin{equation} \label{e49}
\frac{b}{B} = \big(\frac{b}{B}\big)_0
\exp\Big(-2\int_{x_0}^x a \,dx\Big),
\end{equation}
then a simple relation between the second
order ODE for $u$ and the one-dimensional Schr\"odinger equation exists. Hence, many of
the known exact solutions to the Schr\"odinger equation for different potentials can be
utilized to arrive at the solutions to various types of new special
cases of Riccati equation.

When applied to the multidimensional GPE of BECs, the case with
constant $A$ and $B$ yields closed form solutions for the chirp
function $a(t)$ of the matter wave:
\begin{equation} \label{e50}
a(t) = -\frac{s\lambda}{\sqrt{B}}\sqrt{-\frac{\alpha
(t)}{2\beta (t)}} + e^{\phi(t)}\Big[\frac{1}{a_0 +
\frac{s\lambda}{\sqrt{B}}\sqrt{-\frac{\alpha_0}{2\beta_0}}} +2
\int_0^t \beta (\tau) e^{\phi (\tau)} \,d\tau\Big]^{-1},
\end{equation}
given that the following relation
holds between the diffraction coefficient $\beta$ and the strength
of the parabolic potential $\alpha$:
\begin{equation} \label{e51}
\sqrt{-\frac{\beta}{\alpha}} =
\sqrt{-\frac{\beta_0}{\alpha_0}} - \frac{2\sqrt{2}sA}{\sqrt{B}}
\int_0^t \beta \,dt.
\end{equation}
 Here $\phi (t) = -{2\sqrt{2}s\lambda}
 \int_0^t \sqrt{-\alpha (\tau)\beta (\tau)}
\,d\tau/{\sqrt{B}}$.
The chirp function is an essential
part of the exact solutions to GPE.

\subsection*{Acknowledgments}
We acknowledge the support of the Qatar National Research Foundation
project NPRP 25-6-7-2.


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