A Malmquist Type Theorem for a Class of Delay Differential Equations

Abstract

We show that if the following delay differential equation of rational coefficients

$$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}= \frac{\sum _{i=0}^pa_i(z)w^i}{\sum _{j=0}^qb_j(z)w^j} \end{aligned}$$

admits a transcendental entire solution w of hyper-order less than one, then it reduces into a delay differential equation of rational coefficients

$$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{1}{w(z)}\sum _{i=0}^{k+2}A_{i}(z)w^i(z), \end{aligned}$$

which improves some known theorems obtained most recently by Zhang and Huang. Some examples are constructed to show that our results are accurate.

1 Introduction and Main Result

By using Nevanlinna theory, we obtain a main theorem as follows:

Theorem 1.1

Fix three integers k, n, s with $k\ge 0,n\ge 1, s\ge 1$ and take nonzero complex numbers $c_1,\ldots ,c_s$. Let $a, e_1,\ldots ,e_s$ be rational functions on $\mathbb {C}$ such that $|e_1|+\cdots +|e_s|\ne 0$ and let $R(z,w)(\not \equiv 0)$ be an irreducible rational function of w with rational coefficients on z. If the following delay, differential equation

$$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=R(z,w(z)) \end{aligned}$$

(1.1)

admits a transcendental entire solution w with $\rho _2(w) < 1$, and then, (1.1) reduces into the form:

$$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{1}{w(z)}\sum _{i=0}^{k+2}A_{i}(z)w^i(z), \end{aligned}$$

(1.2)

where $A_i(z)$ are rational functions. In particular, the following two conclusions hold:

• (a) If $a=0$ or w has finitely many zeros when $a\not =0$, we have $A_0=0$.

• (b) If w has infinitely many zeros and there exists a zero $z_0$ of w such that neither a zero of $w^{(n)}$ nor a zero or a pole of $a,e_1,\ldots ,e_s$ and the coefficients in $\frac{P(z,w)}{ Q(z,w)}$ when $a\not =0$, we have $A_0\not =0$.

In Theorem 1.1, the symbol $\rho _2(w)$ denotes the hyper order of w defined by

$$\begin{aligned} \rho _2(w)=\mathop {\overline{\lim }}\limits _{r\rightarrow \infty }\frac{\log \log T(r,w)}{\log r}. \end{aligned}$$

Many researchers had discussed the properties of meromorphic solutions on delay differential equations (or complex difference–differential equations named by some complex analysts). For example, see [3, 4, 6, 14, 16, 17]. Especially, a large number of results on Malmquist-type theorems had been obtained (see [5, 8, 10, 11, 15, 19, 20]).

Next, we recall some special cases of Theorem 1.1. According to the classical Malmquist’s theorem (cf. [1]), if the differential equation

$$\begin{aligned} w'(z)=R(z,w(z)) \end{aligned}$$

(1.3)

has a transcendental meromorphic solution w, then (1.3) reduces into a Riccati equation. For this special case, there is no restriction $\rho _2(w) < 1$, so that we may think that Theorem 1.1 should be true if the restriction $\rho _2(w) < 1$ is canceled.

In [2], Chen mentioned that if the following functional equation

$$\begin{aligned} w(z+1)+w(z-1)=R(z,w(z)) \end{aligned}$$

(1.4)

has an admissible meromorphic solution w of finite order, either R(z, w) reduced into a polynomial of w of degree $\le 2$ or R(z, w) can be transformed into the forms in Painlevé equations I, II, where the order $\rho (w)$ of w is defined by

$$\begin{aligned} \rho (w)=\mathop {\overline{\lim }}\limits _{r\rightarrow \infty }\frac{\log T(r,w)}{\log r}. \end{aligned}$$

Moreover, Zhang and Huang [20] proved that if the following delay differential equation

$$\begin{aligned} w(z+1)-w(z-1)+a(z)\frac{w'(z)}{w(z)}=R(z,w(z)) \end{aligned}$$

(1.5)

admits a transcendental entire solution w with $\rho _2(w) < 1$, then (1.5) reduces into

$$\begin{aligned} w(z+1)-w(z-1)+a(z)\frac{w'(z)}{w(z)}=\frac{1}{w(z)}\sum _{i=0}^{2}A_{i}(z)w^i(z), \end{aligned}$$

in which $A_0=0$ if $a=0$.

For the special case $n=1, k=0, s=2, e_1=c_1=1, e_2=c_2=-\,1$, it is easy to see that Zhang and Huang proved Theorem 1.1 in [20].

Finally, examples 1.2–1.7 show that the form (1.2) in Theorem 1.1 does exist.

Example 1.2

The following functional equation

$$\begin{aligned} w^k(z)\left\{ zw(z+1)-zw(z+2)\right\} =A(z)w^{k+1}(z)-A(z)w^k(z) \end{aligned}$$

has a transcendental entire solution $w(z)=ze^z+1$, in which $A(z)=e(z+1)-e^2(z+2)$.

Example 1.3

The following delay differential equation

$$\begin{aligned} w^k(z)\sum \limits _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=(2\pi i)^n a(z) \end{aligned}$$

admits a transcendental entire solution $w(z)=e^{2\pi iz}$, where $n\ge 1, k\ge 0$ are integers, $\sum \nolimits _{\mu =1}^s e^{2\pi i c_\mu }e_{\mu }(z)=0$, and $a(z)\ (\not \equiv 0)$ is any rational function.

Example 1.4

The following delay differential equation

$$\begin{aligned} w(z)\{w(z+1)-w(z-1)\}+a(z)\frac{w^{(n)}(z)}{w(z)}=A_1(z)w^2(z)+\frac{z+n-1}{z}a(z) \end{aligned}$$

has a transcendental entire solution $w(z)=ze^z$, where $n\ge 1$ is an integer, $A_1(z)=\frac{e(z+1)-e^{-1}(z-1)}{z}$, and $a(z)\ (\not \equiv 0)$ is any rational function.

Example 1.5

The following delay differential equation

$$\begin{aligned} w^k(z)\{w(z+1)-w(z-1)\}-\frac{1}{(2\pi i)^n}\frac{w^{(n)}(z)}{w(z)}=\frac{2w^{k+1}(z)-w(z)+z}{w(z)} \end{aligned}$$

has a transcendental entire solution $w(z)=e^{2\pi iz}+z$, where $n\ge 2, k\ge 0$ are integers.

Example 1.6

The following delay differential equation

$$\begin{aligned} w^k(z)\sum \limits _{\mu =1}^se_\mu (z) w(z+c_\mu )+\frac{1}{(2\pi i)^n}\frac{w^{(n)}(z)}{w(z)}=\frac{w(z)-1}{w(z)} \end{aligned}$$

has a transcendental entire solution $w(z)=e^{2\pi iz}+1$, where $c_\mu \in \mathbb {Z}-\{0\}, n\ge 1$, $k\ge 0$ are integers, and $\sum \nolimits _{\mu =1}^s e_\mu (z)=0$.

Example 1.7

The following delay differential equation

$$\begin{aligned} w^2(z)w(z+1)+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{e(w^4(z)-w^3(z))+a(z)(w(z)-1)}{w(z)} \end{aligned}$$

has a transcendental entire solution $w(z)=e^z+1$, where $n\ge 1$ is an integer and a(z) is any nonzero rational function.

2 Preliminary

Let $\mathcal {M}(\mathbb {C})$ be the field of meromorphic functions on the complex plane $\mathbb {C}$. We assume that the reader is familiar with the basic results of Nevanlinna theory and standard notation of a meromorphic function $w\in \mathcal {M}(\mathbb {C})$ such as characteristic function T(r, w), proximity function m(r, w), counting function N(r, w), the first main theorem, and so on (see, e.g., [9, 12, 18]). As usual, we use S(r, w) to denote any real functions of growth o(T(r, w)) as $r\rightarrow \infty$ outside of a possible exceptional set of finite logarithmic (or linear) measure. A meromorphic function a(z) on $\mathbb {C}$ is called a small function of w(z) if and only if $T(r,a)=S(r,w)$.

Generally, if $R(z,w)(\not \equiv 0)$ is an irreducible rational function of w over the field $\mathcal {M}(\mathbb {C})$, then there exist irreducible polynomials

$$\begin{aligned} P(z,w)=\sum \limits _{i=0}^pa_i(z)w^i,\ Q(z,w)=\sum \limits _{j=0}^qb_j(z)w^j, \end{aligned}$$

(2.1)

where $a_i(z), b_j(z)$ are meromorphic functions on $\mathbb {C}$, such that

$$\begin{aligned} R(z, w) =\frac{P(z,w)}{Q(z,w)}. \end{aligned}$$

(2.2)

Now we introduce a few of lemmas used in this paper.

Lemma 2.1

[12] If the coefficients $a_i$, $b_j$ of P(z, w) and Q(z, w) defined in (2.1) are small functions of an element $w\in \mathcal {M}(\mathbb {C})$, then the function $R(z):=R(z,w(z))$ defined in (2.2) satisfies an estimate

$$\begin{aligned} T(r,R) = \deg _w(R)T(r,w) + S(r, w), \end{aligned}$$

where $\deg _w(R)=\max \{p, q\}$ is the degree of R(z, w).

Lemma 2.2

[7] Let w be a nonconstant meromorphic function on $\mathbb {C}$ and take $c\in \mathbb {C}-\{0\}$. If $\rho _2(w) < 1$ and $\varepsilon > 0$, then

$$\begin{aligned} m\left( r,\frac{w_c}{w}\right) =o\left( \frac{T(r,w)}{r^{1-\rho _2(w)-\varepsilon }}\right) \end{aligned}$$

holds outside of a set of finite logarithmic measure, where $w_c(z)=w(z+c)$ as usual.

Lemma 2.3

Let w be a transcendental meromorphic solution of a delay differential equation

$$\begin{aligned} U(z,w)\varPhi (z, w) = \varPsi (z,w), \end{aligned}$$

where U(z, w) (resp. $\varPhi (z, w), \varPsi (z, w)$) is a difference (resp. differential–difference) polynomial in w such that the coefficients are small meromorphic functions of w and such that U(z, w) contains just one term of maximal degree in w and its shifts. If $\rho _2(w) < 1$, $\deg _w (\varPsi )\le \deg _w (U)$, then the function $\varPhi (z):=\varPhi (z,w(z))$ satisfies an estimate

$$\begin{aligned} m(r,\varPhi )=S(r,w). \end{aligned}$$

The proof of Lemma 2.3 is similar to the proof of Clunie-type lemma in [13]. In fact, it is sufficient to change the shifts $w^{(n)}(z+c)$ in the equation $U(z,w)\varPhi (z,w)=\varPsi (z,w)$ into the form a(z)w(z), where

$$\begin{aligned} a(z)=\frac{w^{(n)}(z+c)}{w(z)}=\frac{w^{(n)}(z+c)}{w^{(n)}(z)}\cdot \frac{w^{(n)}(z)}{w(z)} \end{aligned}$$

is regarded as a coefficient, and then, one apply Lemma 2.2 and logarithmic derivative lemma to equation $U(z,w)\varPhi (z,w)=\varPsi (z,w)$ as done in [13]. Hence, we omit the proof here.

Lemma 2.4

[18] If meromorphic functions $w_1,\ldots , w_n$ on $\mathbb {C}$ and entire functions $g_1,\ldots , g_n$ on $\mathbb {C}$ satisfy the following conditions:

• 1. $\sum _{j=1}^nw_je^{g_j }= 0;$

• 2. $g_h-g_k$ are not constants for $1 \le h< k \le n$;

• 3. $T(r, w_j) = S(r, e^{g_h-g_k})$ for $1 \le j \le n, 1 \le h < k \le n$,

then we have $w_j= 0\ (j = 1, 2,\ldots , n)$.

Lemma 2.5

If w is a transcendental entire solution of (1.1) with $\rho _2(w) < 1$ such that w has finitely many zeros and $a\not =0$, then (1.1) reduces into the form (1.2) with $A_0=0$.

Proof

Since w(z) has finitely many zeros, by means of Hadamard factorization theorem, w(z) can be written as

$$\begin{aligned} w(z)=H(z)e^{g(z)}, \end{aligned}$$

(2.3)

where H is a nonzero polynomial and g is a nonconstant entire function such that $\rho (g) = \rho _2(e^{g}) = \rho _2(w)< 1$. Substituting (2.3) into (1.1), we obtain

$$\begin{aligned} H^k(z)L(z)e^{(k+1)g(z)}+a(z)\frac{B(z)}{H(z)}=R(z,w(z)), \end{aligned}$$

(2.4)

where B is a differential polynomial of $H,\ldots ,H^{(n)},g',\ldots ,g^{(n)}$ and

$$\begin{aligned} L(z)=\sum _{\mu =1}^se_\mu (z)H(z+c_\mu )e^{g(z+c_\mu )-g(z)}. \end{aligned}$$

If $L(z)\equiv 0$, applying Lemma 2.1 to (2.4), we deduce

$$\begin{aligned} \deg _w(R)T(r,w)+S(r,w)=T\left( r,R\right) =S(r,w), \end{aligned}$$

which implies $\deg _w(R)=0$. Therefore, Eq. (1.1) is of the form

$$\begin{aligned} w^k(z)\sum \limits _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=a_0(z), \end{aligned}$$

where $a_0(z)\ (\not \equiv 0)$ is a rational function, so that Lemma 2.5 is proved.

Next, we assume that $L(z)\not \equiv 0$. Since $\rho _2(e^{g})<1$, we apply Lemma 2.2 and obtain

$$\begin{aligned} T(r, L) = m(r, L)+O(\log r) \le \sum _{\mu =1}^s m\left( r,\frac{e^{g_{c_\mu }}}{e^{g}}\right) + O(\log r) = S(r, e^g), \end{aligned}$$

where $g_{c_\mu }(z)=g(z+c_\mu )$. Thus, by using (2.4) and noting that

$$\begin{aligned} T(r,w)=T(r,e^g)+O(\log r), \end{aligned}$$

we find

$$\begin{aligned} \deg _w(R)T(r,w)+S(r,w)=T(r,R)\le (k+1)T(r,e^g)+S(r,e^g), \end{aligned}$$

which means $\deg _w(R)\le k+1$.

If we write R(z, w) into the forms (2.1) and (2.2), where $a_i$, $b_j$ are rational functions, then (2.4) can be reduced into the form

$$\begin{aligned} L\sum _{j=0}^{q}b_{j}H^{j+k}e^{(j+k+1)g}+ \frac{aB}{H}\sum _{j=0}^{q}b_{j}H^{j}e^{j g} -\sum _{i=0}^{p}a_{i}H^{i}e^{i g}=0. \end{aligned}$$

(2.5)

Without loss of generality, we may assume $\deg _w(P)\le k+1, \deg _w(Q)\le k+1$, so that (2.5) becomes

$$\begin{aligned} 0= & {} L\sum _{j=1}^{k+1}b_{j}H^{j+k}e^{(j+k+1)g}+\left( Lb_{0}+aBb_{k+1}-Ha_{k+1}\right) H^{k}e^{(k+1)g} \\&+\sum _{j=0}^{k}\left( \frac{aB}{H}b_{j}-a_{j}\right) H^{j}e^{j g}. \end{aligned}$$

By using Lemma 2.4, we obtain

$$\begin{aligned}&Lb_{j}H^{j+k}= 0, \quad j=1,2,\ldots ,k+1,\\&Lb_{0}+aBb_{k+1}-Ha_{k+1}=0,\\&\frac{aB}{H}b_{j}-a_{j}= 0,\quad j=0,\ldots ,k, \end{aligned}$$

which means

$$\begin{aligned}&b_{k+1}=\cdots =b_1= 0,\quad b_0\not = 0,\\&a_k=\cdots =a_1= 0,\quad a_{k+1}\not = 0,\quad a_0\not = 0. \end{aligned}$$

Therefore, Eq. (1.1) is of the form

$$\begin{aligned} w^k(z)\sum \limits _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{a_{k+1}(z)}{b_0(z)}w^{k+1}(z)+\frac{a_{0}(z)}{b_0(z)}. \end{aligned}$$

Thus, Lemma 2.5 is proved. $\square$

3 Proof of Theorem 1.1

Suppose that w is a transcendental entire solutions of (1.1). Now by multiplying both sides with w and applying division algorithm, we can write (1.1) into the form

$$\begin{aligned} w^{k+1}(z)\sum _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)w^{(n)}(z)=P_1(z,w(z))+\frac{P_2(z,w(z))}{Q(z,w(z))}, \end{aligned}$$

(3.1)

where $\deg _w P_2(z,w)=p_2<q=\deg _w Q(z,w)$ with $P_1=0,\ P_2=Pw$ if $p+1=\deg (Pw)<q$. First of all, we claim that $P_2= 0$. Otherwise, if $P_2\not = 0$, we can rewrite (3.1) into the following form

$$\begin{aligned} Q(z,w)\varPhi (z,w)=P_2(z,w), \end{aligned}$$

(3.2)

where

$$\begin{aligned} \varPhi (z,w(z))=w^{k+1}(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)w^{(n)}(z)-P_1(z,w). \end{aligned}$$

Applying Lemma 2.3 into (3.2), then the function $\varPhi (z):=\varPhi (z,w(z))$ satisfies an estimate

$$\begin{aligned} m(r,\varPhi )=S(r,w), \end{aligned}$$

and hence,

$$\begin{aligned} T(r,\varPhi )=m(r,\varPhi )+N(r,\varPhi )=S(r,w), \end{aligned}$$

(3.3)

since $N(r,\varPhi )=O(\log r)=S(r,w)$. Thus, it follows from (3.2) and (3.3) that

$$\begin{aligned} T(r,Q)\le T(r,P_2)+T(r,\varPhi )+O(1)=T(r,P_2)+S(r,w), \end{aligned}$$

where $Q(z)=Q(z,w(z)), P_2(z)=P_2(z,w(z))$, which combining with Lemma 2.1 yield

$$\begin{aligned} qT(r,w)\le p_2T(r,w)+S(r,w), \end{aligned}$$

that is, $q\le p_2$. This is a contradiction.

Therefore, Eq. (3.1) has the form

$$\begin{aligned} w^{k+1}(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)w^{(n)}(z)=P_1(z,w(z)). \end{aligned}$$

By Lemmas 2.1 and 2.2, we see that the function $P_1(z)=P_1(z,w(z))$ satisfies

$$\begin{aligned} \begin{array}{rl} \deg (P_1)T(r,w)&{}=T\left( r,P_1\right) =m\left( r,w^{k+1}\sum _{\mu =1}^se_\mu w_{c_\mu }+aw^{(n)}\right) \\ &{}\le m\left( r,w\right) +m\left( r,w^{k+1}\right) +m\left( r,\frac{\sum \limits _{\mu =1}^se_\mu w_{c_\mu }}{w}\right) +m\left( r,\frac{aw^{(n)}}{w}\right) \\ &{}\le (k+2)T(r,w)+S(r,w),\\ \end{array} \end{aligned}$$

where $w_{c_\mu }(z)=w(z+c_\mu )$, that is, $\deg (P_1)\le k+2$. Further, (3.1) becomes

$$\begin{aligned} w^{k+1}(z)\sum _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)w^{(n)}(z)=\sum _{i=0}^{k+2}a_iw^i(z) \end{aligned}$$

or equivalently,

$$\begin{aligned} w^{k}(z)\sum _{\mu =1}^se_\mu (z)w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{\sum _{i=0}^{k+2}a_iw^i(z)}{w(z)}. \end{aligned}$$

Thus, we have proved that (1.1) reduced into the form (1.2).

• (a) If $a=0$, by applying division algorithm, we can write (1.1) as below

  $$\begin{aligned} w^k(z)\sum \limits _{\mu =1}^se_\mu (z)w(z+c_\mu )=\hat{P_1}(z,w(z))+\frac{\hat{P_2}(z,w(z))}{Q(z,w(z))}, \end{aligned}$$

  (3.4)

  where $\deg _w \hat{P_2}(z,w)=\hat{p_2}<q=\deg _w Q(z,w)$ with $\hat{P_1}=0, \hat{P_2}=P$ if $p=\deg (P)<q$. Using the method similar to the above, we obtain $\hat{P_2}= 0$. Therefore, Eq. (3.4) has the form

  $$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )=\hat{P_1}(z,w(z)). \end{aligned}$$

  By Lemmas 2.1 and 2.2, we see that the function $\hat{P_1}(z)=\hat{P_1}(z,w(z))$ satisfies

  $$\begin{aligned} \begin{array}{rl} \deg (\hat{P_1})T(r,w)&{}=T\left( r,\hat{P_1}\right) =T\left( r,w^k\sum _{\mu =1}^se_\mu w_{c_\mu }\right) \\ &{}\le T\left( r,w^k\right) +T\left( r,\sum \limits _{\mu =1}^se_\mu w_{c_\mu }\right) \le (k+1)T(r,w)+S(r,w),\\ \end{array} \end{aligned}$$

  where $w_{c_\mu }(z)=w(z+c_\mu )$, which implies $\deg (\hat{P_1})\le k+1$, so that we proved the case $a=0$. If $a\not =0$ and w(z) has finitely many zeros, then Lemma 2.5 shows that Eq. (1.1) is of the form (1.2) with $A_0= 0$.

• (b) Now we assume that w(z) has infinitely many zeros and $a\not =0$. If $W=\sum _{j=1}^se_\mu w_{c_\mu }\not =0$, by applying Lemma 2.1 to (1.1) again, we have

  $$\begin{aligned} \begin{array}{rl} T\left( r,R\right) &{}=T\left( r,w^kW+a\frac{w^{(n)}}{w}\right) \le T\left( r,w^kW\right) +T\left( r,\frac{aw^{(n)}}{w}\right) +O(1)\\ &{}\le (k+2)T(r,w)+S(r,w),\\ \end{array} \end{aligned}$$

  that is,

  $$\begin{aligned} \deg _w(R)T(r,w)\le (k+2)T(r,w)+S(r,w), \end{aligned}$$

  which implies $\deg _w(R)\le k+2$. We write R(z, w) into the forms (2.1) and (2.2) with $\max \{p,q\}\le k+2$, so that (1.1) has the form

  $$\begin{aligned} w^k(z)W+a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{\sum _{i=0}^{k+2}a_i(z)w^i(z)}{\sum _{j=0}^{k+2}b_j(z)w^j(z)}, \end{aligned}$$

  (3.5)

Since w(z) has infinitely many zeros, by the assumption that we can choose a zero $z_0$ of w(z) such that neither a zero of $w^{(n)}(z)$ nor a zero or a pole of $a(z),e_1(z),\ldots ,e_s(z)$ and the coefficients in $\frac{P(z,w)}{ Q(z,w)}$. Now we claim $b_0=0$. Otherwise, if $b_0\not =0$, then $R(z_0,w(z_0)$ is a finite value, but $z_0$ is a pole of the function at left-hand side of (3.5). This is a contradiction. Note that $a_0\not = 0$ since R(z, w) is irreducible in w. Thus, (3.5) becomes

$$\begin{aligned} w^k(z)\sum _{\mu =1}^se_\mu (z) w(z+c_\mu )+a(z)\frac{w^{(n)}(z)}{w(z)} =\frac{\sum _{i=0}^{k+2}a_i(z)w^i(z)}{\sum _{j=1}^{k+2}b_j(z)w^j(z)}, \end{aligned}$$

(3.6)

or equivalently,

$$\begin{aligned} Ww\sum _{j=1}^{k+2}b_{j}w^{j+k} =\sum _{i=0}^{k+2}a_{i}w^{i+1}-a\sum _{j=1}^{k+2}b_{j}w^{j}w^{(n)}. \end{aligned}$$

Further, we claim

$$\begin{aligned} b_{k+2}=b_{k+1}=\cdots =b_3= 0. \end{aligned}$$

In fact, if there exists some $j\in \{3,\ldots , k+2\}$ with $b_j\not = 0$, then Lemma 2.3 yields two relations immediately

$$\begin{aligned} T\left( r,W\right) =m\left( r,W\right) +O(\log r)=S(r,w), \end{aligned}$$

and

$$\begin{aligned} T\left( r,wW\right) =m\left( r,wW\right) +O(\log r)=S(r,w), \end{aligned}$$

which imply

$$\begin{aligned} T(r,w)=S(r,w). \end{aligned}$$

This is a contradiction. Hence, (3.6) becomes the form

$$\begin{aligned} w^k(z)W(z)+a(z)\frac{w^{(n)}(z)}{w(z)} =\frac{\sum _{i=0}^{k+2}a_{i}(z)w^{i}(z)}{b_2(z)w^2(z)+b_1(z)w(z)}. \end{aligned}$$

(3.7)

Moreover, we claim $b_2=0$. In fact, if $b_2\not =0$, we multiply both sides of (3.7) by $b_2(z)w^2(z)+b_1(z)w(z)$ to get

$$\begin{aligned} b_2w^{k+2}W=\sum _{i=0}^{k+2}a_{i}w^{i}-b_1w^{k+1}W-b_2aww^{(n)}-ab_1w^{(n)}. \end{aligned}$$

Applying Lemma 2.3 to the above equation, we obtain

$$\begin{aligned} m\left( r,W\right) =S(r,w), \end{aligned}$$

which yields

$$\begin{aligned} T\left( r,W\right) =m\left( r,W\right) +O(\log r)=S(r,w). \end{aligned}$$

(3.8)

On the one hand, we have

$$\begin{aligned} \begin{aligned} (k+2)T(r,w)+S(r,w)&=T\left( r,\frac{a_{k+2}w^{k+2}+\cdots +a_0}{b_2w^2+b_1w}\right) \\&=T\left( r,w^kW+a\frac{w^{(n)}}{w}\right) \\&\le T\left( r,W\right) +T\left( r,w^k\right) +T\left( r,\frac{aw^{(n)}}{w}\right) \\&\le T\left( r,W\right) +kT(r,w)+N\left( r,\frac{1}{w}\right) +S(r,w)\\&\le T\left( r,W\right) +(k+1)T(r,w)+S(r,w), \end{aligned} \end{aligned}$$

that is,

$$\begin{aligned} T(r,w)\le T\left( r,W\right) +S(r,w)=S(r,w). \end{aligned}$$

This is a contradiction. Therefore, $b_2=0$. Thus, (3.7) becomes

$$\begin{aligned} w^k\sum _{\mu =1}^se_\mu w_{c_\mu }+a\frac{w^{(n)}}{w}=\frac{\sum _{i=0}^{k+2}a_{i}(z)w^{i}(z)}{b_1(z)w(z)}. \end{aligned}$$

If $W=\sum _{j=1}^se_\mu w_{c_\mu }=0$, by applying Lemma 2.1 to (1.1), we have

$$\begin{aligned} \begin{array}{rl} T\left( r,R\right) &{}=T\left( r,a\frac{w^{(n)}}{w}\right) = m\left( r,\frac{aw^{(n)}}{w}\right) +N\left( r,\frac{1}{w}\right) \\ &{}\le T(r,w)+S(r,w),\\ \end{array} \end{aligned}$$

which implies $\deg _w(R)\le 1$. Therefore, Eq. (1.1) is of the form

$$\begin{aligned} a(z)\frac{w^{(n)}(z)}{w(z)}=\frac{a_1(z)w(z)+a_0(z)}{b_1(z)w(z)+b_0(z)}. \end{aligned}$$

(3.9)

Since w(z) has infinitely many zeros, and by a similar reasoning as above, we obtain $b_0=0$ and $a_0\not = 0$. Thus, (1.1) becomes

$$\begin{aligned} w^k\sum _{\mu =1}^se_\mu w_{c_\mu }+a\frac{w^{(n)}}{w}=\frac{a_1(z)w(z)+a_0(z)}{b_1(z)w(z)}, \end{aligned}$$

so that the proof of conclusion (b) is completed.

4 A Note on Theorem 1.1

If the coefficients in (1.1) are constants, we may describe a rational solution of (1.1) as follows:

Theorem 4.1

Fix a nonnegative integer k. If the equation

$$\begin{aligned} e_1w^kw_{c_1}+a\frac{w'}{w}=\frac{\sum _{i=0}^pa_iw^i}{\sum _{j=0}^qb_jw^j} \end{aligned}$$

(4.1)

of constant coefficients has a rational solution $w=M/T$ in which M, T are irreducible polynomials. Then, we have either $\deg (M)=\deg (T)$; or $\deg (M)=\deg (T)-1$, $a_0=0,\ b_0\not =0$; or $\deg (M)>\deg (T), p=q+k+1$.

Examples 4.2 and 4.3 show that all cases in Theorem 4.1 can happen.

Example 4.2

The polynomial $w(z)=z+1$ is a solution of the equation

$$\begin{aligned} w(z)w(z-2)-2\frac{w'(z)}{w(z)}=\frac{w^3(z)-2w^2(z)-2}{w(z)}, \end{aligned}$$

where $e_1=1, c_1=a=-\,2$, $3=p=q+k+1=3$ and $1=m>t=0$.

Example 4.3

The rational function $w(z)=\frac{1}{z}$ solves the equation

$$\begin{aligned} w(z+1)+\frac{w'(z)}{w(z)}=\frac{-w^2(z)}{w(z)+1}, \end{aligned}$$

where $e_1=c_1=a=1$, $m+1=t$ and $a_0=a_1=0, b_0=1$.

Proof

Substituting $w=\frac{M}{T}$ into (4.1), we obtain

$$\begin{aligned} \frac{e_1M^kM_{c_1}}{T^kT_{c_1}}+ a\frac{M'T-MT'}{MT} =\frac{\sum _{i=0}^pa_iM^{i}T^{p-i+q}}{\sum _{j=0}^qb_jM^{j}T^{q-j+p}}, \end{aligned}$$

that is,

$$\begin{aligned} P_1=Q_1+Q_2-Q_3, \end{aligned}$$

(4.2)

where

$$\begin{aligned} P_1&=\sum _{i=0}^pa_iM^{i+1}T^{p-i+k+1+q}T_{c_1},\quad Q_1=e_1\sum _{j=0}^qb_jM^{j+k+1}T^{q-j+p+1}M_{c_1},\\ Q_2&=a\sum _{j=0}^qb_jM^{j}T^{q-j+p+k+1}M'T_{c_1},\quad Q_3=a\sum _{j=0}^qb_jM^{j+1}T^{q-j+p+k}T'T_{c_1}. \end{aligned}$$

Set $m:=\deg (M)$ and $t:=\deg (T)$. When $\deg (M)\ne \deg (T)$, we distinguish two cases to prove Theorem 4.1.

Case 1. $m<t$.

Now we claim $a_0=0, b_0\not =0$. Note that $(a_0,b_0)\not =(0,0)$ since M and T are irreducible. If $a_0\not =0, b_0\not =0$, we have

$$\begin{aligned} \deg (P_1)&= m+(q+p+k+2)t,\quad \deg (Q_1)=(k+2)m+(q+p+1)t,\\ \deg (Q_2)&=\deg (Q_3)=m+(k+2+p+q)t-1<\deg (P_1). \end{aligned}$$

Therefore, we have $\deg (P_1)=\deg (Q_1)$, that is,

$$\begin{aligned} (k+2)m+(q+p+1)t=m+(q+p+k+2)t \end{aligned}$$

which yields $m=t$. This is a contradiction.

If $a_0\not =0, b_0=0$, then there exists some integer $\iota$ with $1\le \iota \le t$ such that $b_0=\cdots =b_{\iota -1}=0$. Thus, we have

$$\begin{aligned} \deg (P_1)&= m+(q+p+k+2)t,\quad \deg (Q_1)=(k+2+\iota )m+(q+p-\iota )t,\\ \deg (Q_2)&=\deg (Q_3)=(\iota +1)m+(k+2+p+q-\iota )t-1, \end{aligned}$$

which means

$$\begin{aligned} \deg (Q_1) \le \deg (Q_2)=\deg (Q_3) < \deg (P_1). \end{aligned}$$

This is a contradiction. Hence, the claim is proved, that is, $b_0\not =0, a_0=0$.

Further, if $a_1\not =0$, we have

$$\begin{aligned} \deg (P_1)&= 2m+(q+p+k+1)t,\quad \deg (Q_1)=(k+2)m+(q+p+1)t,\\ \deg (Q_2)&=\deg (Q_3)=m+(k+2+p+q)t-1, \end{aligned}$$

which imply

$$\begin{aligned} \deg (Q_1)\le \deg (P_1)\le \deg (Q_2)=\deg (Q_3), \end{aligned}$$

and $\deg (Q_2-Q_3)=\deg (Q_2)$ since $m<t$. Thus we have $\deg (P_1)=\deg (Q_2)$, that is,

$$\begin{aligned} 2m+(q+p+k+1)t=m+(k+2+p+q)t-1, \end{aligned}$$

which yields $m=t-1$.

Moreover, there exists some integer $\gamma$ with $2\le \gamma \le m$ such that $a_0=\cdots =a_{\gamma -1}=0, a_{\gamma }\not =0$. Now we find

$$\begin{aligned} \deg (P_1)&= (\gamma +1)m+(q+p+k+2-\gamma )t,\\ \deg (Q_1)&=(k+2)m+(q+p+1)t,\\ \deg (Q_2)&=\deg (Q_3)=m+(k+2+p+q)t-1. \end{aligned}$$

Note that $\deg (P_1)<\deg (Q_2)$ and $\deg (Q_2-Q_3)=\deg (Q_2)$ since $m<t$. We obtain $\deg (Q_1)=\deg (Q_2)$, that is,

$$\begin{aligned} (k+2)m+(q+p+1)t=m+(k+2+p+q)t-1, \end{aligned}$$

which means

$$\begin{aligned} m=t-1/(k+1). \end{aligned}$$

It follows that $m=t-1$ since $k (\ge 0), m, t$ are integers.

Case 2. $m> t$.

Now we find

$$\begin{aligned} \deg (P_1)&= (p+1)m+(q+k+2)t,\quad \deg (Q_1)=(q+k+2)m+(p+1)t,\\ \deg (Q_2)&=\deg (Q_3)=(q+1)m+(k+2+p)t-1<\deg (Q_1). \end{aligned}$$

Thus, it follows that $\deg (P_1)=\deg (Q_1)$, that is,

$$\begin{aligned} (q+k+2)m+(p+1)t=(p+1)m+(q+k+2)t, \end{aligned}$$

which implies $p=q+k+1$. Hence, Theorem 4.1 is proved. $\square$