3.1 The proof of Theorem 1.1
Suppose that (1.1) admits a nontrivial meromorphic solution such that , then by Lemma 2.1, for any given ϵ such that , we have
(3.1)
for all r outside of a possible exceptional set with finite logarithmic measure.
Applying Lemma 2.2, we have
(3.2)
(3.3)
for all r outside of a possible exceptional set with finite linear measure.
Therefore from (1.1) we can get that
i.e.,
(3.4)
Setting , ().
Case 1. , which is .
Subcase 1.1. Assume that . By Lemma 2.4, for the above ϵ, there is a ray such that (where and are defined as in Lemma 2.4, is of linear measure zero) satisfying , or , for a sufficiently large r.
Since , (≢0) () are entire functions and , then when , for a sufficiently large r, by Lemma 2.3, we have
(3.5)
(3.6)
By (3.5) and (3.6), we have
(3.7)
Since , we know that , then . Therefore, by (3.2) we obtain
(3.8)
Substituting (3.1), (3.2), (3.3), (3.7) and (3.8) into (3.4), we obtain
(3.9)
By , and , we know that (3.9) is a contradiction.
When , , using a proof similar to the above, we can get a contradiction.
Subcase 1.2. Assume that . By Lemma 2.4, for the above ϵ, there is a ray such that (where and are defined as in Lemma 2.4, is of linear measure zero) satisfying .
Since , , and , then , thus . For a sufficiently large r, by Lemma 2.3, we get
(3.10)
(3.11)
By (3.10) and (3.11), we get
(3.12)
where .
Since , we see that , then , .
Since , we know that , then . Therefore, by (3.2) we obtain
(3.13)
Substituting (3.1), (3.2), (3.3), (3.12) and (3.13) into (3.3), we obtain
(3.14)
By , and , we know that (3.14) is a contradiction.
Case 2. , which is .
Subcase 2.1. Assume that , then . By Lemma 2.4, for the above ϵ, there is a ray such that (where and are defined as in Lemma 2.4, is of linear measure zero) satisfying . Since , we have . For a sufficiently large r, by Lemma 2.3, we get
(3.15)
(3.16)
By (3.15) and (3.16), we get
(3.17)
Using the same reasoning as in Subcase 1.1, we can get a contradiction.
Subcase 2.2. Assume that , then . By Lemma 2.4, for the above ϵ, there is a ray such that (where and are defined as in Lemma 2.4, is of linear measure zero), then , , .
Since , , and , then , thus . For a sufficiently large r, we get (3.10), (3.11) and (3.12) hold.
Using the same reasoning as in Subcase 1.2, we can get a contradiction. Thus we have .
3.2 The proof of Theorem 1.2
Since , (≢0) () are meromorphic functions and , then when , for a sufficiently large r, by Lemma 2.5, we have
(3.18)
(3.19)
Then, using a similar argument to that of Theorem 1.1, we obtain a contradiction.
3.3 The proof of Theorems 1.3 and 1.4
Suppose that is a nonconstant meromorphic solution of (1.1) such that . We first prove that . Submitting into (1.1), we get
where
Since , we have .
Now, for any given , applying Lemma 2.1 and Theorem 1.1, we can deduce that
This implies that
then follows.
Next, we assert that either or . If the assertion does not hold, we have .
Assume that is a zero (or a pole) of of order m. Applying the Hadamard factorization of a meromorphic function, we write as follows:
where , are entire functions such that , and is a polynomial such that deg .
Now, we obtain from (1.1) that
(3.20)
where
and
Notice that deg for and . Thus, Lemma 2.6 is valid for (3.20), hence we get that for , a contradiction to our assumption. This completes our proof.
The proof of Theorem 1.4 is similar to that of Theorem 1.3.
3.4 The proof of Theorem 1.5
Let f (≢0) be a meromorphic solution of (1.4). Suppose that , then by Lemma 2.1, for any given , there exists a subset of finite logarithmic measure such that for all z satisfying , and as r sufficiently large, we have
(3.21)
Set , and (). Then is a set of linear measure zero.
Consider that , (≢0) are meromorphic functions and , by Lemma 2.5, for the above , there exists a set of linear measure zero such that for any satisfying , and as , we have
-
(i)
if , then
(3.22)
-
(ii)
if , then
(3.23)
Set , then is a set of linear measure zero.
Since are distinct complex constants, then there exists only one such that for any . Now we take a ray such that .
Let , , then . We discuss the following two cases.
Case 1. . We rewrite (1.4) in the form
(3.24)
By (3.21), (3.22) and (3.24), we get for and sufficiently large ,
(3.25)
When , by (3.25), we get
This is impossible.
Case 2. . By (3.21), (3.23) and (3.24), we get for and sufficiently large ,
This is a contradiction. Hence we get .
3.5 The proof of Theorem 1.6
Considering (), (≢0), (≢0) are meromorphic functions and , , by Lemmas 2.2 and 2.5, we know that for any given , there exists a subset of finite logarithmic measure such that, for all z satisfying , and as , we have
-
(i)
if , then
(3.26)
-
(ii)
if , then
(3.27)
Then, using a similar argument to that of Theorem 1.5 and Theorem 1.1 and only replacing (3.22) (or (3.23)) by (3.26) (or (3.27)), we can prove Theorem 1.6.
3.6 The proof of Theorem 1.7
Let be a meromorphic solution of (1.5). Suppose that , then by Lemma 2.7, we obtain . This contradicts , therefore we have .
Suppose that there exist two distinct meromorphic solutions , of Eq. (1.5) such that , then is a meromorphic solution of the homogeneous linear difference equation corresponding to (1.5) and . By Theorem 1.5 or Theorem 1.6, we get a contradiction. So Eq. (1.5) has at most one meromorphic solution satisfying .
Next we prove in the case . Suppose that , then by the Weierstrass factorization, we obtain
(3.28)
where , are entire functions such that , and is a polynomial of degree 1.
In the case . Substituting (3.28) into (1.5), we get
(3.29)
Since are distinct complex numbers, by Lemma 2.7, we obtain that the order of the left-hand side of (3.29) is 1. This contradicts .
For , by using an argument similar to the above, we also obtain a contradiction.
It is obvious that provided that . Therefore we have .