Suppose that *f* is an entire solution with *ρ*_{2}(*f*) < 1 to (2). Obviously, *f* is a transcendental function. For the simplicity, we replace *f*(*z*), *f*′(*z*) and *L*(*z*, *f*) by *f*, *f*′ and *L*, respectively.

By differentiating both sides of (2), we obtain

$$\begin{array}{}{\displaystyle 3{f}^{2}{f}^{\prime}+{L}^{\prime}={\alpha}_{1}{p}_{1}{\mathrm{e}}^{{\alpha}_{1}z}+{\alpha}_{2}{p}_{2}{\mathrm{e}}^{{\alpha}_{2}z}.}\end{array}$$(6)

Combining (2) and (6) yields

$$\begin{array}{}{\displaystyle {\alpha}_{2}{f}^{3}+{\alpha}_{2}L-3{f}^{2}{f}^{\prime}-{L}^{\prime}=({\alpha}_{2}-{\alpha}_{1}){p}_{1}{\mathrm{e}}^{{\alpha}_{1}z}.}\end{array}$$(7)

By differentiating (7) again, we derive that

$$\begin{array}{}{\displaystyle 3{\alpha}_{2}{f}^{2}{f}^{\prime}+{\alpha}_{2}{L}^{\prime}-6f({f}^{\prime}{)}^{2}-3{f}^{2}{f}^{\u2033}-{L}^{\u2033}={\alpha}_{1}({\alpha}_{2}-{\alpha}_{1}){p}_{1}{\mathrm{e}}^{{\alpha}_{1}z}.}\end{array}$$(8)

It follows from (7) and (8) that

$$\begin{array}{}{\displaystyle f\phi =T(z,f),}\end{array}$$(9)

where

$$\begin{array}{}{\displaystyle \phi ={\alpha}_{1}{\alpha}_{2}{f}^{2}-3({\alpha}_{1}+{\alpha}_{2})f{f}^{\prime}+6({f}^{\prime}{)}^{2}+3f{f}^{\u2033},}\end{array}$$(10)

$$\begin{array}{}{\displaystyle T(z,f)=-{\alpha}_{1}{\alpha}_{2}L+({\alpha}_{1}+{\alpha}_{2}){L}^{\prime}-{L}^{\u2033}.}\end{array}$$

Two cases will now be considered below, depending on whether or not *φ* vanishes identically.

If *φ* ≡ 0, then (9) shows that *T*(*z*, *f*) ≡ 0, namely

$$\begin{array}{}{\displaystyle {L}^{\u2033}-({\alpha}_{1}+{\alpha}_{2}){L}^{\prime}+{\alpha}_{1}{\alpha}_{2}L=0.}\end{array}$$(11)

Further, the general solution of (11) is given by

$$\begin{array}{}{\displaystyle L={c}_{1}{\mathrm{e}}^{{\alpha}_{1}z}+{c}_{2}{\mathrm{e}}^{{\alpha}_{2}z},}\end{array}$$(12)

where *c*_{1} and *c*_{2} are constants. Thus, (2) and (12) would give

$$\begin{array}{}{\displaystyle {f}^{3}=({p}_{1}-{c}_{1}){\mathrm{e}}^{{\alpha}_{1}z}+({p}_{2}-{c}_{2}){\mathrm{e}}^{{\alpha}_{2}z}.}\end{array}$$(13)

We claim that *p*_{1} = *c*_{1} or *p*_{2} = *c*_{2}. Assume now, contrary on the assertion, that *p*_{1} ≠ *c*_{1} and *p*_{2} ≠ *c*_{2}. We rewrite (13) as

$$\begin{array}{}{\displaystyle (\frac{{\mathrm{e}}^{-{\alpha}_{2}z/3}f}{\sqrt[3]{{p}_{2}-{c}_{2}}}{)}^{3}-(\sqrt[3]{\frac{{p}_{1}-{c}_{1}}{{p}_{2}-{c}_{2}}}{\mathrm{e}}^{({\alpha}_{1}-{\alpha}_{2})z/3}{)}^{3}=1,}\end{array}$$

which contradicts Lemma 2.1. Hence, *p*_{1} = *c*_{1} or *p*_{2} = *c*_{2}. In this case, we can derive the conclusions (2) and (3).

In the following, we will consider the case *φ* ≢ 0. In this case, (9) gives

$$\begin{array}{}{\displaystyle \phi (z)=\frac{T(z,f)}{f(z)}.}\end{array}$$(14)

Since *T*(*z*, *f*) is a difference-differential polynomial in *f* of degree 1, and *L*(*z*, 0) ≡ 0, it follows from (14), Lemma 2.4 and the lemma on the logarithmic derivatives that *m*(*r*, *φ*) = *S*(*r*, *f*). Note that *φ* is an entire function, so *T*(*r*,*φ*) = *m*(*r*, *φ*) = *S*(*r*, *f*), which means that *φ* is a small function of *f*.

Now, we rewrite (10) as

$$\begin{array}{}{\displaystyle \frac{\phi}{{f}^{2}}={\alpha}_{1}{\alpha}_{2}-3({\alpha}_{1}+{\alpha}_{2})\frac{{f}^{\prime}}{f}+6(\frac{{f}^{\prime}}{f}{)}^{2}+3\frac{{f}^{\u2033}}{f}.}\end{array}$$(15)

Applying the lemma on the logarithmic derivatives to (15), we find *m*(*r*,
$\begin{array}{}{\displaystyle \frac{\phi}{{f}^{2}}}\end{array}$) = *S*(*r*, *f*). Since *φ* is a small function of *f*, one can get *m*(*r*,
$\begin{array}{}{\displaystyle \frac{1}{f}}\end{array}$) = *S*(*r*, *f*). Thus, the first fundamental theorem implies *T*(*r*, *f*) = *N*(*r*,
$\begin{array}{}{\displaystyle \frac{1}{f}}\end{array}$) + *S*(*r*, *f*).

On the other hand, by (10) again, we have *N*_{(2}(*r*, 1/*f*) = *S*(*r*, *f*), and

$$\begin{array}{}{\displaystyle T(r,f)={N}_{1)}(r,\frac{1}{f})+S(r,f),}\end{array}$$(16)

where *N*_{1)}(*r*,
$\begin{array}{}{\displaystyle \frac{1}{f}}\end{array}$) denotes the counting function corresponding to simple zeros of *f*. Differentiating (10) yields

$$\begin{array}{}{\displaystyle {\phi}^{\prime}=2{\alpha}_{1}{\alpha}_{2}f{f}^{\prime}-3({\alpha}_{1}+{\alpha}_{2})[f{f}^{\u2033}+({f}^{\prime}{)}^{2}]+15{f}^{\prime}{f}^{\u2033}+3f{f}^{\u2034}.}\end{array}$$(17)

For brevity, in the following, we assume that *z*_{0} is a simple zero of *f*, and we can assume, without loss of generality, by (16) that *ψ*(*z*_{0}) ≠ 0, ∞, where *ψ* is any non-vanishing small function of *f*. Thus, (10) enables us to deduce the following fact

$$\begin{array}{}{\displaystyle \phi ({z}_{0})=6({f}^{\prime}({z}_{0}){)}^{2}.}\end{array}$$(18)

Now, we are ready to present *φ*′ ≢ 0. Suppose, contrary to our assertion, that *φ*′ ≡ 0, namely, *φ* is a constant, say *A*.

If *z*_{0} is a zero of *f*′(*z*) –
$\begin{array}{}{\displaystyle \sqrt{A/6},}\end{array}$ then we set

$$\begin{array}{}{\displaystyle h(z)=\frac{{f}^{\prime}(z)-\sqrt{A/6}}{f(z)}.}\end{array}$$(19)

Trivially, *h* ≢ 0. Then by the lemma on the logarithmic derivatives, the facts *m*(*r*,1/*f*) = *S*(*r*, *f*), *N*_{(2}(*r*,1/*f*) = *S*(*r*, *f*) and (18), we have *T*(*r*, *h*) = *S*(*r*, *f*).

By (19), we therefore have

$$\begin{array}{}{\displaystyle {f}^{\prime}=hf+\sqrt{A/6},\phantom{\rule{1em}{0ex}}{f}^{\u2033}=({h}^{\prime}+{h}^{2})f+h\sqrt{\frac{A}{6}}.}\end{array}$$(20)

Substituting (20) into (10) yields

$$\begin{array}{}{\displaystyle [{\alpha}_{1}{\alpha}_{2}-3({\alpha}_{1}+{\alpha}_{2})h+3{h}^{\prime}+9{h}^{2}]f=3[({\alpha}_{1}+{\alpha}_{2})-5h]\sqrt{\frac{A}{6}},}\end{array}$$

which implies

$$\begin{array}{}{\displaystyle {\alpha}_{1}+{\alpha}_{2}\equiv 5h,\phantom{\rule{1em}{0ex}}{\alpha}_{1}{\alpha}_{2}-3({\alpha}_{1}+{\alpha}_{2})h+9{h}^{2}\equiv 0.}\end{array}$$

Thereby we have

$$\begin{array}{}{\displaystyle h=\frac{{\alpha}_{1}+{\alpha}_{2}}{5}=\frac{{\alpha}_{1}}{3},\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}h=\frac{{\alpha}_{1}+{\alpha}_{2}}{5}=\frac{{\alpha}_{2}}{3}.}\end{array}$$(21)

Thus, (19) and (21) would give

$$\begin{array}{}{\displaystyle f(z)=\frac{B}{h}{\mathrm{e}}^{hz}-\frac{1}{h}\sqrt{\frac{A}{6}},}\end{array}$$(22)

where *B* is a nonzero constant.

On the other hand, substituting (22) into (2), it follows by Lemma 2.5 that
$\begin{array}{}{\displaystyle \frac{1}{h}\sqrt{\frac{A}{6}}=0.}\end{array}$ This, however, contradicts (16), and thus *φ*′ ≢ 0.

Using the same way as above, *φ*′ ≢ 0 is also obtained by setting

$$\begin{array}{}{\displaystyle \hslash (z)=\frac{{f}^{\prime}(z)+\sqrt{A/6}}{f(z)}}\end{array}$$

assuming that *f*′(*z*_{0}) +
$\begin{array}{}{\displaystyle \sqrt{A/6}=0.}\end{array}$

Moreover, (17) gives

$$\begin{array}{}{\displaystyle {\phi}^{\prime}({z}_{0})=[-3({\alpha}_{1}+{\alpha}_{2})({f}^{\prime}{)}^{2}+15{f}^{\prime}{f}^{\u2033}]({z}_{0})=0.}\end{array}$$(23)

In order to prove Theorem 1.6, we discuss two cases below:

Case 1

[2*φ*′ + (*α*_{1} + *α*_{2})*φ*]*f*′ – 5*φ**f*″ ≡ 0.

In this case, let us write it in the following form

$$\begin{array}{}{\displaystyle {f}^{\u2033}=[\frac{2}{5}\frac{{\phi}^{\prime}}{\phi}+\frac{1}{5}({\alpha}_{1}+{\alpha}_{2})]{f}^{\prime}:=s{f}^{\prime},}\end{array}$$(24)

and consequently

$$\begin{array}{}{\displaystyle {f}^{\u2034}=({s}^{\prime}+{s}^{2}){f}^{\prime}.}\end{array}$$(25)

Substituting (24) and (25) into (17), we then immediately derive

$$\begin{array}{}{\displaystyle {\alpha}_{1}{\alpha}_{2}{\phi}^{\prime}f=[2{\alpha}_{1}{\alpha}_{2}\phi -3({\alpha}_{1}+{\alpha}_{2})s\phi +3({s}^{\prime}+{s}^{2})\phi +3({\alpha}_{1}+{\alpha}_{2}){\phi}^{\prime}-3s{\phi}^{\prime}]{f}^{\prime},}\end{array}$$

which is impossible by (16) and the facts that the coefficients *φ*′(≢ 0), 2*α*_{1}*α*_{2}*φ* – 3(*α*_{1} + *α*_{2})*sφ* + 3(*s*′ + *s*^{2})*φ* + 3(*α*_{1} + *α*_{2})*φ*′ – 3*sφ*′ are small functions of *f*. So, this case can not occur.

Case 2

[2*φ*′ + (*α*_{1} + *α*_{2})*φ*]*f*′ – 5*φf*″ ≢ 0.

Obviously, in this case, by (18), (23) and *f*′(*z*_{0}) ≠ 0, we then see that *z*_{0} is a zero of the function [2*φ*′ + (*α*_{1} + *α*_{2})*φ*]*f*′ – 5*φf*″.

Accordingly, we set

$$\begin{array}{}{\displaystyle \varphi (z)=\frac{[2{\phi}^{\prime}(z)+({\alpha}_{1}+{\alpha}_{2})\phi (z)]{f}^{\prime}(z)-5\phi {f}^{\u2033}(z)}{f(z)}.}\end{array}$$(26)

Then by the lemma on the logarithmic derivatives, the facts *N*_{(2}(*r*, 1/*f*) = *S*(*r*, *f*), *m*(*r*, 1/*f*) = *S*(*r*, *f*), and (17), we have *T*(*r*, *ϕ*) = *S*(*r*, *f*). Thereby, from (26), we obtain

$$\begin{array}{}{\displaystyle {f}^{\u2033}=[\frac{2}{5}\frac{{\phi}^{\prime}}{\phi}+\frac{1}{5}({\alpha}_{1}+{\alpha}_{2})]{f}^{\prime}-\frac{\varphi}{5\phi}f:=s{f}^{\prime}+tf.}\end{array}$$(27)

Trivially, *s*, *t* are small functions of *f*.

By (10) and (27), we have

$$\begin{array}{}{\displaystyle a{f}^{2}+bf{f}^{\prime}+6({f}^{\prime}{)}^{2}=\phi ,}\end{array}$$(28)

where *a* = *α*_{1}*α*_{2} + 3*t*, *b* = 3[*s* – (*α*_{1} + *α*_{2})].

In the following, we consider two subcases.

Subcase 2.1

Suppose that *a* ≡ 0. In this case, (28) becomes

$$\begin{array}{}{\displaystyle (bf+6{f}^{\prime}){f}^{\prime}=\phi ,}\end{array}$$

which gives

$$\begin{array}{}{\displaystyle {f}^{\prime}={\phi}_{1}{\mathrm{e}}^{\beta},\phantom{\rule{1em}{0ex}}bf+6{f}^{\prime}={\phi}_{2}{\mathrm{e}}^{-\beta},}\end{array}$$(29)

where *β*, *φ*_{1} and *φ*_{2} are entire functions such that *φ*_{1}*φ*_{2} = *φ*.

Trivially, in this case, *b* ≢ 0, and it follows by (29) that

$$\begin{array}{}{\displaystyle f=\frac{{\phi}_{2}{\mathrm{e}}^{-\beta}-6{\phi}_{1}{\mathrm{e}}^{\beta}}{b}.}\end{array}$$(30)

Thus, by (29) and (30), we have

$$\begin{array}{}{\displaystyle [{\phi}_{1}+6(\frac{{\phi}_{1}}{b}{)}^{\prime}+6\frac{{\phi}_{1}}{b}{\beta}^{\prime}]{\mathrm{e}}^{2\beta}=(\frac{{\phi}_{2}}{b}{)}^{\prime}-\frac{{\phi}_{2}}{b}{\beta}^{\prime},}\end{array}$$

which shows that

$$\begin{array}{}{\displaystyle (\frac{{\phi}_{2}}{b}{)}^{\prime}-\frac{{\phi}_{2}}{b}{\beta}^{\prime}\equiv 0.}\end{array}$$(31)

Obviously, (31) gives log
$\begin{array}{}{\displaystyle \frac{{\phi}_{2}}{b}=\beta +C,}\end{array}$ where *C* is a constant. Therefore, e^{β} is a small function of *f*, this shows that *f*′ is also a small function of *f*. The contradiction *T*(*r*, *f*) = *S*(*r*, *f*) now follows by (30).

Subcase 2.2

*a* ≢ 0. In this case, applying Lemma 2.7 to (28), we immediately get the following equation

$$\begin{array}{}{\displaystyle 6({b}^{2}-24a)\frac{{\phi}^{\prime}}{\phi}+b({b}^{2}-24a)-6({b}^{2}-24a{)}^{\prime}\equiv 0.}\end{array}$$(32)

Now, we consider two cases.

Firstly, assume that *b*^{2} – 24*a* ≢ 0.

Note that
$\begin{array}{}{\displaystyle b=\frac{6}{5}\frac{{\phi}^{\prime}}{\phi}-\frac{12}{5}({\alpha}_{1}+{\alpha}_{2}),}\end{array}$ and we then rewrite (32) as

$$\begin{array}{}{\displaystyle \frac{11}{5}\frac{{\phi}^{\prime}}{\phi}-\frac{12}{5}({\alpha}_{1}+{\alpha}_{2})=\frac{({b}^{2}-24a{)}^{\prime}}{{b}^{2}-24a}.}\end{array}$$

By integrating the above equation, we have

$$\begin{array}{}{\displaystyle 11\mathrm{log}\phi -12({\alpha}_{1}+{\alpha}_{2})z=5\mathrm{log}({b}^{2}-24a)+\mathrm{log}D,}\end{array}$$(33)

where *D* is a constant. Obviously, (33) gives *φ*^{11} = *D*(*b*^{2} – 24*a*)^{5}e^{12(α1+α2)z}. If *α*_{1} + *α*_{2} ≠ 0, then e^{α1z}, e^{α2z} are also two small functions of *f* because *φ* and *b*^{2} – 24*a* are small functions of *f*. Rewrite (2) as

$$\begin{array}{}{\displaystyle {f}^{2}=-\frac{L}{f}+\frac{{p}_{1}{\mathrm{e}}^{{\alpha}_{1}z}+{p}_{2}{\mathrm{e}}^{{\alpha}_{2}z}}{f}.}\end{array}$$

Then

$$\begin{array}{}{\displaystyle 2m(r,f)=m(r,{f}^{2})=m(r,-\frac{L}{f}+\frac{{p}_{1}{\mathrm{e}}^{{\alpha}_{1}z}+{p}_{2}{\mathrm{e}}^{{\alpha}_{2}z}}{f})}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\le m(r,\frac{1}{f})+S(r,f)=S(r,f),}\end{array}$$

which is a contradiction, since *f* is an entire function. Therefore, *α*_{1} + *α*_{2} = 0.

Combining (2) and (6) yields

$$\begin{array}{}{\displaystyle {\alpha}_{1}{f}^{3}+{\alpha}_{1}L-3{f}^{2}{f}^{\prime}-{L}^{\prime}=({\alpha}_{1}-{\alpha}_{2}){p}_{2}{\mathrm{e}}^{{\alpha}_{2}z}.}\end{array}$$(34)

It follows by (7), (34) and *α*_{1} + *α*_{2} = 0 that

$$\begin{array}{}{\displaystyle {f}^{4}[-{\alpha}_{1}^{2}{f}^{2}+9({f}^{\prime}{)}^{2}]=2{\alpha}_{1}^{2}L{f}^{3}-6{L}^{\prime}{f}^{2}{f}^{\prime}+({\alpha}_{1}L{)}^{2}-({L}^{\prime}{)}^{2}-4{\alpha}_{1}^{2}{p}_{1}{p}_{2}.}\end{array}$$(35)

Obviously, *P*_{4}(*f*) := 2
$\begin{array}{}{\displaystyle {\alpha}_{1}^{2}}\end{array}$*Lf*^{3} – 6*L*′*f*^{2}*f*′ + (*α*_{1}*L*)^{2} – (*L*′)^{2} – 4
$\begin{array}{}{\displaystyle {\alpha}_{1}^{2}}\end{array}$*p*_{1}*p*_{2} is a difference-differential polynomial of *f*, and its total degree at most 4.

If *P*_{4}(*f*) ≡ 0, it follows from (35) that 9(*f*′)^{2} –
$\begin{array}{}{\displaystyle {\alpha}_{1}^{2}}\end{array}$*f*^{2} ≡ 0, and then
$\begin{array}{}{\displaystyle {f}^{\prime}=\pm \frac{{\alpha}_{1}}{3}f.}\end{array}$ Substituting the above expression into (10), we arrive at *φ* ≡ 0, a contradiction. Therefore, *P*_{4}(*f*) ≢ 0. Set *β* = 9(*f*′)^{2} –
$\begin{array}{}{\displaystyle {\alpha}_{1}^{2}}\end{array}$*f*^{2}. In this case, we rewrite (35) as

$$\begin{array}{}{\displaystyle \beta =\frac{{P}_{4}(f)}{{f}^{4}},}\end{array}$$

which, Lemma 2.4 and the fact that we have proved *m*(*r*,
$\begin{array}{}{\displaystyle \frac{1}{f}}\end{array}$) = *S*(*r*, *f*) must show that *m*(*r*, *β*) = *S*(*r*, *f*), *i.e*. *β* is a small function of *f*. Moreover, from Lemma 2.8, it is easy to see that *β* is a constant. By differentiating both sides of *β* = 9(*f*′)^{2} –
$\begin{array}{}{\displaystyle {\alpha}_{1}^{2}}\end{array}$*f*^{2}, we get

$$\begin{array}{}{\displaystyle {f}^{\u2033}-(\frac{{\alpha}_{1}}{3}{)}^{2}f=0.}\end{array}$$(36)

It follows from (36) that

$$\begin{array}{}{\displaystyle f(z)={c}_{1}{\mathrm{e}}^{\frac{{\alpha}_{1}}{3}z}+{c}_{2}{\mathrm{e}}^{-\frac{{\alpha}_{1}}{3}z},}\end{array}$$(37)

where *c*_{1}, *c*_{2} are constants. By (37) and (2), we have
$\begin{array}{}{\displaystyle {c}_{1}^{3}={p}_{1},\phantom{\rule{thinmathspace}{0ex}}{c}_{2}^{3}={p}_{2}.}\end{array}$ Conclusion (1) has consequently been proved.

Now, we assume that *b*^{2} – 24*a* ≡ 0.

By making use of (28), we have
$\begin{array}{}{\displaystyle 6({f}^{\prime}+\frac{b}{12}f{)}^{2}=\phi ,}\end{array}$ which shows that
$\begin{array}{}{\displaystyle \gamma :={f}^{\prime}+\frac{b}{12}f}\end{array}$ is a small function of *f*. Thus,
$\begin{array}{}{\displaystyle {f}^{\u2033}=(\frac{{b}^{2}}{144}-\frac{{b}^{\prime}}{12})f+{\gamma}^{\prime}-\frac{b}{12}\gamma .}\end{array}$

Substituting two above expressions into (10), we obtain

$$\begin{array}{}{\displaystyle [{\alpha}_{1}{\alpha}_{2}+\frac{b}{4}({\alpha}_{1}+{\alpha}_{2})+\frac{{b}^{2}}{144}-\frac{{b}^{\prime}}{4}]{f}^{2}+[-3({\alpha}_{1}+{\alpha}_{2})\gamma -b\gamma +3({\gamma}^{\prime}-\frac{{b}^{\prime}}{12}\gamma )]f=\phi .}\end{array}$$(38)

Using (38) and Lemma 2.9, we therefore have

$$\begin{array}{}{\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{\alpha}_{1}{\alpha}_{2}+\frac{b}{4}({\alpha}_{1}+{\alpha}_{2})+\frac{{b}^{2}}{144}-\frac{{b}^{\prime}}{4}\equiv 0,}\\ {\displaystyle -3({\alpha}_{1}+{\alpha}_{2})\gamma -b\gamma +3({\gamma}^{\prime}-\frac{{b}^{\prime}}{12}\gamma )\equiv 0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\phi \equiv 0.}\end{array}$$

This contradicts *φ* ≢ 0.

Thus, we finish the proof of Theorem 1.6.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.