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Solutions of the third order Cauchy difference equation on groups
Advances in Difference Equations volume 2014, Article number: 203 (2014)
Let be a function, where is a group and is an abelian group. In this paper, the following third order Cauchy difference of = − − − − + + + + + + − − − − (), is studied. We first give some special solutions of on free groups. Then sufficient and necessary conditions on finite cyclic groups and symmetric groups are also obtained.
It is well known from  that Jensen’s functional equation
with the additional condition , is equivalent to Cauchy’s equation
on the real line. Let be a group, be an abelian group. Let and denote the identity elements. The study of (1.1) was extended to groups for f maps G into H in , where the general solution for a free group H with two generators and was given, respectively. Later, the results were generalized to all free groups and , (see ). Since functional equations involve Cauchy difference, which made it become much more interesting [4–7]. For a function , its Cauchy difference, , is defined by
The first order Cauchy difference will be abbreviated as Cf. In , by using the reduction formulas and relations, as given in [2, 3], the general solution of the second order Cauchy difference equation was provided on free groups. Particularly, the authors also gave the expression of general solutions on symmetric group and finite cyclic group.
In this paper, we consider the following functional equation:
It follows from (1.4) that (1.5) is equivalent to the vanishing third order Cauchy difference equation
The purpose of this paper is to determine the solutions of (1.5) on some given groups. Clearly, the general solution of (1.5) will be denoted by
Remark 1 (1) is an abelian group under the pointwise addition of functions; (2) .
2 Properties of solution
Lemma 1 Suppose that . Then
for all and .
Proof Putting in (1.5) we get (2.1). Then from (2.1) we obtain (2.2)-(2.3). Furthermore, by the definition of , we have
One can easily check that
Hence, the above relations imply that is a homomorphism. Similarly, the fact is also true for both and . This proves (2.4).
We now consider (2.5). Actually, it is trivial for by (2.1) and the definition of Cf. Suppose that (2.5) holds for all natural numbers smaller than , then
where the definition of and (2.4) are used in the second equation. This gives (2.5) for all . On the other hand, for any fixed integer , by (1.4) and (2.1), we have
from which it follows that
from (2.4) and the above claim for . This confirms (2.5) for . □
Remark 2 For any function , the following statements are pairwise equivalent:
The function ;
is a homomorphism;
is a homomorphism;
is a homomorphism.
Before presenting Proposition 1, we first introduce the following useful lemma, which was given in .
Lemma 2 (Lemma 2.4 in )
The following identity is valid for any function and :
Proposition 1 Suppose that . Then
for and all , such that , .
Proof Replacing in (2.6) by we have
The vanishing of for yields
Therefore, by (2.5) and (2.4), we have
which is (2.7). This completes the proof. □
Remark 3 In particular, if , then Proposition 1 holds.
3 Solution on a free group
In this section, we study the solutions on a free group. We first solve (1.5) for the free group G on a single letter a.
Theorem 1 Let G be the free group on one letter a. Then if and only if it is given by
Proof Necessity. It can be obtained from (2.5) in Lemma 1.
Sufficiency. Taking (3.1) as the definition of f on . By Remark 2, we only need to verify that is a homomorphism with respect to each variable and thus f belongs to . Let
be any three elements of G. Then it follows from (1.4) and (3.1) that
By a tedious calculation, we have
which leads to the result that is a homomorphism with respect to each variable. □
At the end of this section, for the free group on an alphabet with , we discuss some special solutions of (1.5).
An element can be written in the form
For each fixed and fixed pair of distinct , define the functions W, , :
Proposition 2 For any fixed and fixed pair of distinct a, b in , the following assertions hold:
belongs to ;
belongs to ;
belongs to .
Proof Claim (i) follows from the fact that is a morphism from to ℤ by (3.6).
Now we consider assertion (ii). Let x, y, z, w in the free group be written as
Hence, we have
This concludes assertion (ii). Claim (iii) follows from (3.7) directly. □
4 Solution on symmetric group
The symmetric group on a finite set X is the group whose elements are all bijective functions from X to X and whose group operation is that of function composition. If , then it is called the symmetric group of degree n and denoted .
In this section, we consider (1.5) for .
Lemma 3 If , then
for all , , and all rearrangements π.
Proof Note that is a homomorphism and H is an abelian group, which yields
This proves (4.1). By a similar procedure, we can also verify (4.2)-(4.3). □
Lemma 4 Let τ be an arbitrary 2-cycle in and , then
Proof It suffices to prove (4.7). The proofs for the rest of the statements are straightforward. Using , and (1.4) we get
which implies that since . This completes the proof. □
Lemma 5 For any 2-cycle σ, τ, υ, and , we have
Proof For any 2-cycle σ, there exists such that . Hence, for any , by (4.3) we have
In particular, (4.8) follows from (4.9)-(4.11). □
Lemma 6 Assume that for every 2-cycle . Then for any , , rearrangement π, where , β are 2-cycles, we have
for every .
Proof Firstly, for any 2-cycle , and rearrangement π, it follows from the assumption , Proposition 1, and (4.8) that
which gives (4.12).
On the other hand, for every there exist 2-cycles , , , such that , and . Note that for some 2-cycles , we have
by (4.12). In particular, taking in (4.13), we obtain (4.14). This completes the proof. □
According to Lemma 6, we give the following main result in this section.
Theorem 2 Assume that for every 2-cycle . Then if and only if there is an such that and
Proof Necessity. Let . Then for any , there exist 2-cycles , , such that . In view of (4.8), (4.14), (4.5)-(4.6), and Proposition 1, we get
Let , we claim that
We first prove the even case. Obviously, (4.17) is true for since . For an inductive proof, suppose that (4.17) also holds for , . Then we compute that
which yields . This confirms the even case of (4.17). When p is odd, (4.17) is true for because of . Suppose that (4.17) holds for , and then we get
This completes the proof of (4.17). According to (4.17), (4.16) becomes
This proves that f must have the form (4.15) with .
Sufficiency. Let be defined by (4.15), where is a constant with . In order to prove the identity of (1.5), by the symmetry of , , , it suffices to verify the following four cases: case (i) is odd, and , , are even; case (ii) , are odd, and , are even; case (iii) , , are odd, and is even; case (iv) , , and are odd.
In fact, for case (i) it is easy to see that , , , , , , , are odd, and , , , , , , are even, which leads to the equality of (1.5). The proofs of the other cases are similar. □
5 Solution on the finite cyclic group
Let be a cyclic group of order n with generator a. In this section, we study the general solution on the finite cyclic group .
Theorem 3 Assume that n is odd and . Then if and only if it is given by
where and satisfy
Proof Necessity. Let be a function satisfying (1.5). Then by (2.5), we see that f also satisfies (5.1), i.e.,
Let in (5.1), since and is an integer, the summand and by the fact that , , we obtain
Furthermore, by using (2.4), (2.3), and , we get
This proves (5.2)-(5.3).
Sufficiency. We claim that (5.1)-(5.3) defines a function on . Indeed, for each , by (5.1) and (5.3) we have
where the last identity is obtained because , (5.2), and n is odd.
Finally, for any , , , and , we have
which, after a long and tedious computation, gives 0. Consequently, . This completes the proof. □
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The authors are grateful to the editor and the referees for their valuable comments and suggestions, especially for the improvement of Theorem 2. This paper is supported by the National Science Foundation of China (11301226, 61202109), Zhejiang Provincial Natural Science Foundation of China under Grant No. LQ13A010017.
The authors declare that they have no competing interests.
The authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.