Uniqueness for meromorphic solutions of Schwarzian differential equation

1.   Introduction and main results

In this paper, we use the basic notions of Nevanlinna's theory, see, e.g., ^[3,5,10]. In addition, we use the notation

to denote the order of growth of the meromorphic function $f$. Let $S(r, f)$ denote any quantity satisfying $S(r, f) = o\left(T(r, f)\right)$ for all $r$ outside of a set with finite logarithmic measure.

Let $f$ and $g$ be two meromorphic functions, $a$ be a small function relative to both $f$ and $g$. We say that $f$ and $g$ share $a$ CM if $f-a$ and $g-a$ have the same zeros with the same multiplicities. $f$ and $g$ share $a$ IM if $f-a$ and $g-a$ have the same zeros ignoring multiplicities. Nevanlinna's four values theorem (see ^[8,9]) says that if two nonconstant meromorphic functions $f$ and $g$ share four values CM, then $f = g$ or $f$ is a Möbius transformation of $g$. The condition "$f$ and $g$ share four values CM'' has been weakened to "$f$ and $g$ share two values CM and two values IM'' by Gundersen ^[1,2], as well as by Mues ^[7].

For Schwarzian differential equation

Ishizaki [4,Theorem 1] showed that if the Schwarzian Eq (1.1) possesses an admissible solution, then $d+2k\sum\limits_{j = 1}^{l}\delta(\alpha_{j}, f)\leq 4k$, where $\alpha_{j}$ are distinct complex constants, and $d = \deg R(z, f)$. In particular, when $R(z, f)$ is independent of $z$, it is shown that if (1.1) possesses an admissible solution $f$, then by some Möbius transformation $w = (af+b)/(cf+d) (ad-bc\neq 0)$, $R(z, f)$ can be reduced to some special forms, see [4,Theorem 3]. Liao and Ye^[6] considered differential equation, which is a simple type of the Schwarzian differential equation, and gave the order of meromorphic solutions as follows.

Theorem 1. [6,Theorem 3] Let $P$ and $Q$ be polynomials with $\deg P = m$ and $\deg Q = n$ and let $R(z) = P(z)/Q(z)$ and $k$ be a positive integer. If $f$ is a transcendental meromorphic solution of the equation

then $m-n+2k > 0$ and the order $\rho(f) = (m-n+2k)/2k$.

In the follows, we apply the Nevanlinna theory and uniqueness of meromorphic functions to the Schwarzian differential Eq (1.2), and investigate the uniqueness of meromorphic solutions if the meromorphic solution $f$ shares three values with any other meromorphic function. we obtain

Theorem 2. Let $R$ be a nonconstant rational function and $k$ be a positive integer. Suppose that $f$ is a transcendental meromorphic solution of the differential Eq (1.2) without single poles. If a meromorphic function $g$ shares $a, b, \infty$ CM with $f$, then $f\equiv g$.

2.   Lemmas

We now give some preparations.

Lemma 1. [9,Theorem 1.51] Suppose that $n\geq 2$ and let $f_1, \cdots, f_n$ be meromorphic functions and $g_1, \cdots, g_n$ be entire functions such that

(i) $\sum_{j = 1}^nf_j\exp\{g_j\} = 0$;

(ii) when $1\leq j < k\leq n, \, g_j-g_k$ is not constant;

(iii) when $1\leq j\leq n, \, 1\leq h < k\leq n,$

where $E\subset (1, \infty)$ has finite linear measure or logarithmic measure. Then $f_j\equiv 0\; (j = 1, \cdots, n).$

Lemma 2. Let $f$ be a meromorphic solution of Eq (1.2), then $f'$ is a meromorphic solution of equation

where $Q$ is a nonconstant rational function.

Proof. Set

We then prove that $Q$ is a nonconstant rational function.

Since $f$ is of finite order by Theorem 1, (2.1) shows $Q$ is also of finite order and

We see from (1.2) that

where $R_{1}$ is some nonconstant rational function. Thus (2.2) and (2.3) show that

that is,

Since $R_{1}$ is a nonconstant rational function, we deduce from (2.4) that $Q$ cannot be a constant. We then prove $Q$ cannot be transcendental.

By comparing the poles of the both sides of (1.2), we get that $f^{'}$ has only finitely many zeros and poles under the assumption that $f$ without single poles. Thus $Q$ has only finitely many single poles. If $Q$ is transcendental, then

contradicting to (2.4). Hence, $Q$ is a nonconstant rational function.

Lemma 3. Let $a, b$ be two distinct constants, $\beta, \gamma$ be nonconstant polynomials with $\deg\beta \neq \deg \gamma$, and

Then $f$ cannot be a meromorphic solution of Eq (1.2).

Proof. Assume that $f$ is a meromorphic solution of Eq (1.2). Lemma 2 shows

Without loss of generality, we assume $Q$ is a nonconstant polynomial. Otherwise, we just multiply the dominator of $Q$ to both sides of (2.6). We now divide our proof into two cases.

Case 2.1. $\deg \beta > \deg \gamma.$ Rewriting (2.5) as

where

Obviously,

Since $e^{\beta}$ is of regular growth order $\deg\beta$, we see $a_{01}, a_{00}$ are small functions of $e^{\beta}$. We conclude from (2.7) that

where

and

where

We deduce from (2.8), (2.11) and $\rho (a_{01}') = \rho(a_{01})$, $\rho(a_{00}') = \rho(a_{00})$ that

We assert that $a_{11} \not\equiv 0.$ Otherwise, (2.11) shows

Applying the method of separating variables to Eq (2.15), we have $a_{01} = ce^{-\beta}$, thus $\rho(a_{01}) = \deg\beta$, contradicting to (2.8). Similarly, we also get

and $a_{21}\not\equiv 0.$ Substituting (2.10), (2.12) into (2.6), we obtain

where

By (2.14), (2.16), (2.18), we have

Thus, $\rho(A_{1j}) < \deg \beta\; \; (j = 0, 1).$ Since $e^{\beta}$ is of regular growth order $\deg \beta,$ we obtain

Applying Lemma 1 to (2.17), we have

Thus, we obtain from (2.13), (2.18) that

Applying the method of separating variables to the above equations, we have

that is

By (2.7), we also have

Substituting (2.11), (2.21) into (2.20), we obtain

Applying Lemma 1 and $\deg\beta > \deg\gamma$, we have $-c\gamma'\equiv0$, then $c = 0$, thus $a_{11} = 0$, a contradiction.

Case 2.2. $\deg \beta < \deg \gamma.$ Rewriting (2.5) as

where

Obviously,

Thus, we conclude from (2.22) that

where

and

where

We deduce from (2.24), (2.26) that

We assert that $b_{11} \not\equiv 0.$ Otherwise, (2.26) shows

Applying the method of separating variables to Eq (2.30), we have $b_{00} = ce^{\gamma},$ and $\rho(b_{00}) = \deg\gamma.$ contradicting to (2.24). Similarly, we also get

and $b_{22}\not\equiv 0.$

Substituting (2.25), (2.27) into (2.6), we obtain

where

By (2.29), (2.31), we have

Thus, $\rho(A_{2j}) < \deg \gamma\; (j = 0, 1, 2).$ Since $e^{\gamma}$ is of regular growth order $\deg \gamma,$ we obtain

Applying Lemma 1 to (2.32), we have

Thus, we obtain from (2.28), (2.33) that

Applying the method of separating variables to equation, we have

that is

By (2.23), we have

Substituting (2.26), (2.36) into (2.35), we obtain

Applying Lemma 1 and $\deg\beta < \deg\gamma$, we have $c\beta'\equiv0$, then $c = 0$, thus $b_{11} = 0$, a contradiction. In conclusion, $f$ of the form (2.5) cannot be a meromorphic solution of Eq (1.2).

3.   Proof of Theorem 2

Proof. Since $f$ and $g$ share $a, b, \infty$ CM, we can get

By applying the second fundamental theorem to function $g$, we obtain

Similarly, we can get $T(r, f) \leq 3 T(r, g)+S(r, f)$, so $T(r, g) = O(T(r, f)+S(r, f))$, and then $\rho(g) = \rho(f) < \infty$ by Theorem 1.

Furthermore, there exist polynomials $\alpha$ and $\beta$ such that

and

Assume, to the contrary, that $f\not \equiv g$. Then from (3.1) and (3.2), we obtain

Again by (3.1) and (3.2), we get

or

where $\gamma = \beta- \alpha$ is a nonzero polynomial.

We now suppose that $\alpha$ and $\beta$ are not all constants. Otherwise, we easily obtain a contradiction by (3.3). Thus, we split our proofs into two cases.

Case 1. $\alpha$ is a constant and $\beta$ is non-constant polynomial.

Denote $e^{-\alpha} = B(\neq0)$. then $B\not\equiv1$, (3.3) shows

which yields

Substituting (3.6), (3.7) into Eq (2.6), we conclude that

where

By (3.9), we have

Thus, $\rho(A_{3j}) < \deg \gamma (j = 0, 1).$ Since $e^{\beta}$ is of regular growth order $\deg \beta,$ we obtain

Applying Lemma 1 to (3.8), we have

By $A_{30}\equiv 0$, we obtain

Applying the method of separating variables to equation, we have

By (2.6), we have

Substituting (3.6), (3.11) into Eq (3.10), we conclude that

Thus, we have $\beta$ is a constant, a contradiction.

Case 2. $\beta$ is a constant and $\alpha$ is non-constant polynomial. Similar to the proof of Case 1, we also get a contradiction. We deduce from (3.4) and Lemma 3 that $\deg \beta = \deg \gamma,$ and

Substituting (3.13) into Eq (2.6), we conclude that

where

Obviously, we obtain that

Thus,

Therefore, Eq (3.14) can be rewritten as

In the following, we divide our proof into four cases.

Case A. $\deg(\beta+\gamma) < \deg\gamma.$ Combining this with $\deg \beta = \deg \gamma,$ we get $\deg(\beta-\gamma) = \deg \gamma, \deg(\beta - 2\gamma) = \deg \gamma.$ Thus, $e^{\beta}, e^{\gamma}, e^{\beta-\gamma}, e^{\beta-2\gamma}$ are of regular growth order $\deg \gamma$. Equation (3.17) shows that

where

By this and (3.16), we obtain $\rho(B_{00}) \leq \max\{\rho(A_{44}), \rho(A_{40}), \deg(\beta + \gamma)\} < \deg \gamma = \deg\beta.$ Then

Together with (3.18) and Lemma 1, we have

By $A_{42} \equiv 0$ and (3.15), we have

In Case A, we again split two subcases.

Subcase A.1. If $\deg\gamma\geq2$. Applying the method of separating variables to equation, we have

Substituting (3.11), (3.13) into Eq (3.20), we conclude that

By (3.21), $\deg(\beta+ \gamma) < \deg \gamma$ and Lemma 1, we obtain $\gamma'(z)\equiv0$, thus $\gamma(z)$ is a constant, a contradiction.

Subcase A.2. If $\deg\gamma = 1$. Let $\gamma(z) = m z + n_{1},$ where $m \neq 0, n_{1}$ are constants. Hence, $\gamma' = m, \gamma'' = 0.$ Substituting these into Eq (3.19), we get

that is $Q = -m$, a contradiction.

Case B. $\deg(\beta-\gamma) < \deg\gamma.$ Equation (3.17) shows that

Together with (3.16), (3.22), $\deg(\beta-\gamma) < \deg\gamma$ and Lemma 1, we have

Substituting (3.15) and $\beta = \alpha + \gamma$ into the last equality $A_{41}e^{\beta-\gamma}+A_{40}\equiv0$, we have

Together with (3.23) and Lemma 1, we have

In Case B, we again split two subcases.

Subcase B.1. If $\deg\gamma\geq2$. Applying the method of separating variables to equation, we have

Substituting (3.11), (3.13) into Eq (3.25), we conclude that

By (3.26), $\deg(\beta-\gamma) < \deg \gamma$ and Lemma 1, we obtain $\gamma'\equiv0$, a contradiction.

Subcase B.2. If $\deg\gamma = 1$. Let $\gamma = mz + n_{1},$ where $m \neq 0, n_{1}$ are constants. Hence, $\gamma' = m, \gamma'' = 0.$ Substituting these into Eq (3.24), we get

that is $Q = m$, a contradiction.

Case C. $\deg(\beta-2\gamma) < \deg\gamma.$ Equation (3.17) shows that

By $\deg \beta = \deg \gamma$ and $\deg(\beta- 2\gamma) < \deg \gamma$, we have $\deg(\beta- \gamma) = \deg(\beta + \gamma) = \deg \gamma$. By this and (3.16), we have

Together with (3.27) and Lemma 1, we have

By $A_{40}\equiv0$ and (3.15), we also have (3.24). Using the same method as the above Subcase B, we get a contradiction.

Case D. $\deg(\beta+\gamma) = \deg(\beta-\gamma) = \deg(\beta-2\gamma) = \deg\gamma$. By this and (3.15), for $j = 0, 1, 2, 3, 4,$ we have

Combining this with Lemma 1, we have

By $A_{40}\equiv0$ and (3.15), we also have (3.24). Using the same method as the above Subcase B, we get a contradiction.

4.   Conclusions

Together with the Nevanlinna theory and uniqueness of meromorphic functions, this paper considers the certain type of Schwarzian differential equation, and investigate the uniqueness of meromorphic solutions if the meromorphic solution $f$ shares three values with any other meromorphic function.

Acknowledgments

We would like to thank the referee for his/her thorough reviewing with constructive suggestions and comments to improve the present paper. This research was supported partly by the National Natural Science Foundation of China (11801093, 11871260) and Characteristic Innovation Project of Guangdong Province (2019KTSCX119).

Conflict of interest

The authors declare no conflict of interest.