Rational solutions of a Riccati equation

We consider the problem of solutions of the equation

$$\begin{equation} f'+f^2-\sum_{j=-2}^2 a_j x^j=0 \end{equation} \tag{ 1 }$$

that are rational in the variable $x$. It is easy to see that such a solution must have the form

$$\begin{equation} f=k_1 x+k_0+\frac{k_{-1}}{x}+\sum_{i=1}^{m}\frac{1}{x+y_{i}}\,. \end{equation} \tag{ 2 }$$

Proposition 1.  A function (2) solves (1) if and only if

$$\begin{equation} \begin{gathered} \, a_2=k_1^2, \quad a_1=2 k_0 k_1,\quad a_0=k_0^2+2 k_1 k_{-1}+(2 m+1) k_1, \\ a_{-1}=2 k_0k_{-1}+2 m k_0-2 k_1 \sum_{i=1}^{m} y_{i},\quad a_{-2}=k_{-1} (k_{-1}-1) \end{gathered} \end{equation} \tag{ 3 }$$

and

$$\begin{equation} k_0-k_1 y_i-\frac{k_{-1}}{y_i}=\sum_{j\ne i}\frac{1}{y_i-y_{j}}, \qquad i=1,\dots,m. \end{equation} \tag{ 4 }$$

The coefficients $a_i$ in (1) are defined up to the scaling $x\mapsto \mu x$, $f\mapsto \mu^{-1} f$. For definiteness, in what follows we will assume that $a_2=1$, so that $k_1^2=1$. It is easy to see that, as $x\to \infty$, a rational solution of (1) has an expansion

$$\begin{equation} f^{+}=x+\frac{1}{2}a_1+\frac{1}{8}\,\frac{d-4}{x}+ \sum_{j=2}^\infty f_j^{+}x^{-j}\quad\text{or}\quad f^{-}=-x-\frac{1}{2}a_1-\frac{1}{8}\,\frac{d+4}{x}+ \sum_{j=2}^\infty f_j^{-} x^{-j}, \end{equation} \tag{ 5 }$$

where $d=4a_0-a_1^2$. The properties of formal solutions of the form (5) and questions related to their convergence were considered in [1].

Example 1.  Assume that the potential in the Riccati equation is an even function (so that $a_1=a_{-1}=0$) and a rational solution is an odd function. Then a solution of the form

$$\begin{equation*} f^{+}=x+\frac{\lambda}{x}+\frac{P'}{P}\,, \qquad P=\sum_{k=0}^{n} a_k x^{2 k}, \end{equation*}$$

exists if and only if the Riccati equation has the form

$$\begin{equation} f'+f^2=x^2+2\lambda+4 n+1+\frac{\lambda(\lambda-1)}{x^2}\,, \end{equation} \tag{ 6 }$$

and the polynomial $P$ solves the linear equation $xP''+2(x^2+\lambda)P'-4nxP=0$. From it we find that $a_k=C^k_n\displaystyle\prod_{i=k+1}^n\biggl(\lambda+i-\dfrac{1}{2}\biggr)$ and $a_n=1$, where $\lambda$ is an arbitrary parameter.

For each positive integer $m$ the equation

$$\begin{equation*} f'+f^2=x^2-2\lambda-4m-1+\frac{\lambda(\lambda-1)}{x^2} \end{equation*}$$

has the solution

$$\begin{equation*} f^{-}=-x+\frac{\lambda}{x}+\frac{Q'}{Q}\,, \quad \text{where } Q=\sum_{k=0}^{m} b_k x^{2 k},\ b_k=C^k_m(-1)^{m-k}\prod_{i=k+1}^m\biggl(\lambda+i-\frac{1}{2}\biggr),\ b_m=1. \end{equation*}$$

Here $\lambda$ is an arbitrary parameter. If $\lambda=-(2n+2m+1)/2$, then (6) has two rational solutions, $f^{+}$ and $f^{-}$.

Theorem.  Assume that (1) with $a_2\ne 0$ has two rational solutions,

$$\begin{equation} f=k_1x+k_0+\frac{k_{-1}}{x}+\sum_{i=1}^{m}\frac{1}{x+y_{i}} \quad\textit{and}\quad \overline f=\overline k_1 x+\overline k_0+\frac{\overline k_{-1}}{x}+ \sum_{i=1}^{\overline m}\frac{1}{x+\overline y_{i}}\,. \end{equation} \tag{ 7 }$$

Then

$$\begin{equation*} \overline k_1=-k_1, \quad \overline k_0=-k_0, \quad k_{-1}=\overline k_{-1}=-\frac{m+\overline m+1}{2}\,, \quad a_{-2}=\frac{(m+\overline m+2)^2-1}{4}\,. \end{equation*}$$

Corollary.  Equation (1) with $a_2\ne 0$ has at most two rational solutions.

Taking $a_2=1$, we can assume without loss of generality that $k_1=1$ and ${\overline k_1=-1}$. Then for fixed $m$ and $\overline m$ and for the $a_{-2}$, $k_1$, $\overline k_1$, $k_{-1}$, and $\overline k_{-1}$ defined above, the relations (3) and (4) take us to a system of $m+\overline m+4$ equations with respect to the $m+\overline m+4$ unknowns $k_0$, $a_1$, $a_0$, $a_{-1}$, $y_i$, and $\overline y_i$. This system can easily be solved for small $m$ and $\overline m$. In particular, the case when $m=\overline m=0$ is described in Example 1.

Example 2.  For $m=1$ and $\overline m=0$ we obtain the Riccati equation

$$\begin{equation*} f'+f^2=\frac{1}{4}x^2+\frac{1}{2}x+\frac{3}{4}+\frac{1}{x}+\frac{2}{x^2}\,, \end{equation*}$$

which has the two rational solutions

$$\begin{equation*} f^{+}=\frac{1}{2}x+\frac{1}{2}-\frac{1}{x}+\frac{1}{x-1}\quad\text{and}\quad f^{-}=-\frac{1}{2}x-\frac{1}{2}-\frac{1}{x}\,. \end{equation*}$$

It is clear that a rational solution $f$ of a Riccati equation defines the solution $\psi=\exp\displaystyle\int f\,dx$ of the corresponding linear Schrödinger equation.

An algorithm for finding elementary solutions of a second-order linear equation with rational coefficients was put forward in [2]. In our case this algorithm is based on the observation that (for fixed numerical values of the coefficients $a_i$ in (1)) the formulae (3) determine at most four possible values of $m$, each of which must be investigated separately. The two rational solutions $f^{+}$ and $f^{-}$ yield [3] the solution $\phi=(f^{+}-f^{-})^{-1}$ of a third-order linear equation which is well known in soliton theory.

The authors acknowledge useful discussions with V. G. Marikhin and are grateful to the participants of the seminar on mathematical physics at the Landau Institute for Theoretical Physics for their interest in this work.