Microsoft Word - Math031007.doc SQU Journal For Science, 9 (2004) 25-31 © 2004 Sultan Qaboos University 25 Solution of a Nonlinear Functional Equation Valery C. Covachev* and H. Ali Yurtsever** *Department of Mathematics and Statistics, College of Science, Sultan Qaboos University, P. O. Box 36, Al Khod 123, Muscat, Sultanate of Oman, Email: valery@squ.edu.om, **Department of Mathematics, Faculty of Science and Literature, Fatih University, Istanbul, Turkey حل معادلة دالية غير خطية فاليري كريستوف كوفاتشف و حسن علي يورتسيفر يعتمد. معادلة دالية غير خطيةبحل الخاصة و) 2004( لنظرية ريستيسكي تعميم هذا البحث نقدم في :خالصة مثل هذا النوع من لحل طريقة المصفوفات مع Λ االختيارية المجموعة في λ علي تعريف البارامتر البرهان .المعادالت ABSTRACT: In the present paper a generalization of a theorem of I.B. Risteski (2004) concerning the solution of a nonlinear functional equation is given. The proof is based on a parametric approach by introducing a parameter λ in an arbitrary set Λ , and on a matrix method for solving linear functional equations. KEYWORDS: Nonlinear functional equation; Symmetric group. 1. Introduction unctional equations find applications in biology, social sciences, engineering, as well as in many other branches of mathematics and a great number of such applications can be found in the monograph by Aczel (1966). This has led to considerable interest in the study of functional equations and has given rise to numerous articles and monographs on this subject. The present paper is devoted to the study of a nonlinear functional equation which generalizes the quadratic complex vector functional equation solved by I.B. Risteski (2004). To the best of our knowledge, up to now this type of nonlinear functional equation has not been considered in the literature. We carried out our research to shed some light on this field of nonlinear functional equations. The results presented here and in (Risteski, 2004) supplement and extend some of the results in (Risteski et al 1999, 2000a, 2000b). In order to prepare the background for our study, we present the result of I.B. Risteski (2004). Let V be a complex finite-dimensional vector space and let there exist mappings , : nf g V V→ . ( ) 1 2i i n≤ ≤Z F VALERY C. COVACHEV and H. ALI YURTSEVER 26 will denote vectors in the vector space V . Multiplication of two arbitrary vectors ( )1 2, , , T nu u u=U … and ( )1 2, , , T nv v v=V … in V is defined as ( )1 1 2 2, , , T n nu v u v u v=UV … . I.B. Risteski (2004) gave the following result. Theorem 1. The general solution of the nonlinear complex vector functional equation ( ) ( )1 1 1 2 2, , , , , ,n n n n nf f− + +Z Z Z Z Z Z… … ( ) ( )1 1 1 2 2, , , , , ,n n n n nf f− + += Z Z Z Z Z Z… … ( ) ( )1 1 2 1 3 2, , , , , , ,n n n n n nf f− + + ++ Z Z Z Z Z Z Z… … ( ) ( )1 1 2 1 2 1, , , , ,n n n n nf f− + −+ + Z Z , Z Z Z Z… … ( )2n ≥ (1) is given by ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 2 1 2 2 2 1 2 1 2 , , , n n n n n n n F F F F F F f F F F = U U U U U U U U U U U U … , (2) where ( ) 1iF i n≤ ≤ are arbitrary functions in V . In the next section we shall consider a slightly generalized version of (1). In the proof of the theorem we shall use techniques developed in Risteski (2002) and Risteski and Covachev (2000, 2001). On the other hand, the method used in Risteski (2004) cannot be used in the proof of this theorem without imposing a very restrictive assumption on the equation considered. 2. Statement of the Problem In this part of our paper, we shall consider the generalized nonlinear complex vector functional equation ( ) ( )1 1 1 2 2, , , , , ,n n n n nf g− + +Z Z Z Z Z Z… … ( ) ( )1 1 1 2 2, , , , , ,n n n n nf g− + += Z Z Z Z Z Z… … ( ) ( )1 1 2 1 3 2, , , , , , ,n n n n n nf g− + + ++ Z Z Z Z Z Z Z… … ( ) ( )1 1 2 1 2 1, , , , ,n n n n nf g− + −+ + Z Z , Z Z Z Z… … ( )2n ≥ . (3) If f g≡ , equation (3) is reduced to (1). It is easy to see that if a component of f is identically 0 , then the corresponding component of g may be arbitrary. Similarly, if a component of g is identically 0 , then the corresponding component of f may be arbitrary. So we need to consider only solutions ( ),f g of equation (3) for which no component of f or g is identically 0 . Thus we may suppose that (3) is a scalar functional equation. Moreover, the arguments ( ) 1 2i i n≤ ≤Z may belong to an arbitrary set V (with at least 1n + distinct elements). Furthermore, it is easily seen that the function f depends on the arguments ( ) 11 1, , nn V −− ∈Z Z… as on a parameter. So SOLUTION OF A NONLINEAR FUNCTIONAL EQUATION 27 we can slightly generalize equation (3) by introducing a parameter λ in an arbitrary set Λ instead of ( ) 11 1, , nn V −− ∈Z Z… . For convenience we write iZ instead of ( ) 0n i i n+ ≤ ≤Z . Thus we consider the equation ( ) ( ) ( ) ( )0 1 2 1 0 2, , , , , , , ,n nf g f gλ λ=Z Z Z Z Z Z Z Z… … ( ) ( ) ( ) ( )2 1 0 3 1 1 0, , , , , , , , ,n n nf g f gλ λ −+ + +Z Z Z Z Z Z Z Z Z… … ( )2 ,n ≥ (4) where : f VΛ× → C and : ng V → C . 3. Main Result Theorem 2. Each solution ( ),f g of the nonlinear functional equation (4), such that none of the functions f and g is identically 0 , is given by ( ) ( ) ( ) 1 , , n i i i f F Gλ λ = = ∑U U (5) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 2 1 2 2 2 1 2 1 2 , , , n n n n n n n G G G G G G g G G G = U U U U U U U U U U U U … , (6) where ( ): : 1i iF and G V i nΛ → → ≤ ≤C C are arbitrary functions in V . Proof : Let us introduce the function ( ) ( ) ( )0 1 0 1, , , , , , , .n nF f gλ λ=Z Z Z Z Z Z… … (7) It satisfies the linear functional equation ( ) ( )0 1 2 1 0 2, , , , , , , , , ,n nF Fλ λ=Z Z Z Z Z Z Z Z… … (8) ( ) ( )2 1 0 3 1 1 0, , , , , , , , , , , .n n nF Fλ λ −+ + +Z Z Z Z Z Z Z Z Z… … We will not solve equation (8). Instead, we will just find the form of the solutions of (8) which can be represented as (7). To this end, we use a method similar to that in Risteski (2002) and in Risteski and Covachev (2001). Below we will denote by nS the symmetric group of degree n (or symmetric group on n letters, see Feyzioğlu (1990)). This is the group of all one-to-one mappings of the set { }1, 2, , n… onto itself, and its elements are called permutations (of 1, 2, , n… ). Here, for our convenience, we denote by 1nS + the group of all one-to-one mappings of the set { }0,1, 2, , n… onto itself. Let 1nSα +∈ . This means that ( ) ( ) ( )( )0 , 1 , , nα α α… is a permutation of the indices ( )0,1, , n… . Then equation (8) is equivalent to the linear homogeneous system ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( )( )0 1 2 1 0 2, , , , , , , , , ,n nF Fα α α α α α α αλ λ−Z Z Z Z Z Z Z Z… … (9) VALERY C. COVACHEV and H. ALI YURTSEVER 28 ( ) ( ) ( ) ( ) ( )( )2 1 0 3, , , , , , nF α α α α αλ− −Z Z Z Z Z… ( ) ( ) ( ) ( )( ) 11 1 0, , , , , 0, .nn nF Sα α α αλ α +−− = ∈Z Z Z Z… This system of ( )1 !n + equations has a nontrivial solution if and only if its determinant is 0 . We will not show that this determinant is 0 , but we will use the form of the nontrivial solution and eventually obtain a solution of (8) admitting the factorization (7). According to Risteski (2002) and Risteski and Covachev (2001), a possible nontrivial solution of (8) has the form ( ) ( ) ( ) ( )( ) 1 0 1 0 1, , , , , , , , , n n n S F C Hα α α α α λ λ +∈ = ∑Z Z Z Z Z Z… … (10) where 1: nH V +Λ× → C is an arbitrary function, and Cα are complex constants (which may also depend on λ ). However, it is easy to see that if ( )0 1, , , , nF λ Z Z Z… has the form (7), then ( )0 1, , , , nH λ Z Z Z… must admit the factorization ( ) ( ) ( ) ( ) ( )0 1 0 0 1 1, , , , ,n n nH C G G Gλ λ=Z Z Z Z Z Z… (11) where : C Λ → C and ( ): 0iG V i n→ ≤ ≤C are arbitrary functions. Thus the representation (10) takes the form ( ) ( ) ( )( ) ( )( ) ( )( ) 1 0 1 0 10 1, , , , , n n n n S F C G G Gα α α α α λ λ +∈ = ∑Z Z Z Z Z Z… (12) where we have denoted ( ) ( )C C Cα αλ λ= . By virtue of (7) we must have ( ) ( ) ( ) ( )( ) ( )( ) ( )( ) 1 0 1 0 10 1, , , . n n n n S f g C G G Gα α α α α λ λ +∈ = ∑Z Z Z Z Z Z… (13) Since g is not identically 0 , there exists an n -tuple ( )1 2, , , nn V∈A A A… such that ( )1 2, , , 0ng ≠A A A… . We set ( )1, ,i i i n= =Z A … in (13) and find ( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 1 0 1 00 1 , , , , n n n ii S i f g C G Gα α α α λ λ − − +∈ = = ∑ ∏Z A A Z A… (14) i.e., ( )0,f λ Z can be represented as ( ) ( ) ( )0 0 0 , . n i i i f D Gλ λ = = ∑Z Z (15) Now equation (13) can be written in the form ( ) ( ) ( ) ( ) ( ) ( ) ( )1 1 0 1 0 0 0 1 , , . n nn n i i n i jj i i S j D G g G C Gα α α λ λ − += = ∈ =   =     ∑ ∑ ∑ ∏Z Z Z Z Z… (16) From (16) it follows that SOLUTION OF A NONLINEAR FUNCTIONAL EQUATION 29 ( ) ( ) ( ) ( ) ( )1 1 1 2 1 , , , , 0,1, , . n n n jj S ji C g G i n D α α α λ λ − +∈ = = =∑ ∏Z Z Z Z… … (17) Now the quotients ( ) ( )/ iC Dα λ λ must all be independent of λ . Moreover, if we compare two of these representations of ( )1 2, , , ng Z Z Z… , for 0i = and for some 0i ≠ , and we give suitable values to all but one of the variables 1 2, , , nZ Z Z… , we find that ( )0G Z is expressed as a linear combination of ( ) ( )1 , , nG GZ Z… . We can then write ( ) ( ) ( )0 0 1 , , n i i i f F Gλ λ = = ∑Z Z (18) ( ) ( )( )1 2 1 , , , . n n n j j S j g c Gα α α∈ = = ∑ ∏Z Z Z Z… (19) Equation (18) shows that, in fact, equation (5) is valid. It remains to prove that (up to a constant factor) equation (6) also holds. To this end we will determine the constants cα (up to a common factor) so that the pair ( ),f g given by (18) and (19) satisfies equation (4): ( ) ( ) ( )( )0 1 1n nn i i j j i S j F G c Gα α α λ = ∈ = ∑ ∑ ∏Z Z ( ) ( ) ( ) ( ) ( )( ) ( ) 1 1 1 01 1 1n n i i j j i S j F G c G Gα αα α α λ − −= ∈ ≠ = ∑ ∑ ∏Z Z Z ( ) ( ) ( ) ( ) ( )( ) ( ) 1 1 2 02 1 2n n i i j j i S j F G c G Gα αα α α λ − −= ∈ ≠ + +∑ ∑ ∏Z Z Z ( ) ( ) ( ) ( ) ( )( ) ( ) 1 1 0 1 . n n i i n j jn i S j n F G c G Gα αα α α λ − −= ∈ ≠ +∑ ∑ ∏Z Z Z Thus for each 1, 2, ,i n= … we must have ( ) ( )( )0 1n n i j j S j G c Gα α α∈ = ∑ ∏Z Z ( ) ( ) ( ) ( )( ) ( ) 1 1 1 01 1n i j j S j G c G Gα αα α α − −∈ ≠ = ∑ ∏Z Z Z ( ) ( ) ( ) ( )( ) ( ) 1 1 2 02 2n i j j S j G c G Gα αα α α − −∈ ≠ + +∑ ∏Z Z Z ( ) ( ) ( ) ( )( ) ( ) 1 1 0 . n i n j jn S j n G c G Gα αα α α − −∈ ≠ + ∑ ∏Z Z Z (20) It is easily seen that each term on the left cancels with some term on the right-hand side of (20). Indeed, let us take the term ( ) ( )( ) ( )( )0i i ji j j i G c G Gα α α ≠ ∏Z Z Z VALERY C. COVACHEV and H. ALI YURTSEVER 30 for some fixed nSα ∈ . Suppose that ( )i kα = . Then this term is identical with the term on the right ( ) ( ) ( ) ( )( ) ( ) 1 1 0 .i k j jk j k G c G Gα αα α − −≠ ∏Z Z Z Now we have to determine (up to a common factor) the coefficients cα so that the remaining terms on the right-hand side of (20) cancel pairwise. The terms on the right ( ) ( ) ( ) ( )( ) ( ) 1 1 0i k j jk j k G c G Gα αα α − −≠ ∏Z Z Z and ( ) ( ) ( ) ( )( ) ( ) 1 1 0i j j j G c G Gβ ββ β − −≠ ∏Z Z Z for k ≠ cancel each other if and only if ( ) ( ) ( ) ( )1 1 , , ,k i k iα β β α− −= = = ( ) ( ) ( )1 , 0.j j j k c cα βα β α −= ∀ ≠ + = This means that ( ) ( ) ( ) { } { }1 1 1, , 1, 2, , \ , .k k m m m n kβα βα βα− − −= = = ∀ ∈ … Thus the permutation 1βα − is a transposition (exchanges the indices k and and keeps all other indices fixed), i.e., β and α are permutations of different signature (one even and one odd). Consequently, we must have if is even, if is odd, c c cα α α  =  − for some constant 0c ≠ . Thus ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 1 2 1 2 1 2 2 2 1 2 1 2 , , , . n n n n n n n G G G G G G g c G G G = Z Z Z Z Z Z Z Z Z Z Z Z … Finally, we can replace the functions ( )iF λ by ( )icF λ for 1, 2, ,i n= … . Thus we have shown that any solution ( ),f g of equation (4), such that none of the functions f and g is identically 0 , must have the form (5) and (6). It remains to show that every pair of functions ( ),f g of the form (5) and (6) is a solution of equation (4). Indeed, by virtue of the equations (5) and (6) we obtain ( ) ( ) ( ) ( )0 1 2 1 0 2, , , , , , , ,n nf g f gλ λ−Z Z Z Z Z Z Z Z… … ( ) ( ) ( ) ( )2 1 0 3 1 1 0, , , , , , , , ,n n nf g f gλ λ −− − −Z Z Z Z Z Z Z Z Z… … SOLUTION OF A NONLINEAR FUNCTIONAL EQUATION 31 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 1 1 0 1 1 1 1 0 1 0 1 0. i i i n n n i i i i i n n n n n G G G G G G F G G G G G G λ = = =∑ Z Z Z Z Z Z Z Z Z Z Z Z This completes the proof of Theorem 2. Remark 3. It is clear that the solutions of equations (1) and (4) as given, respectively, by (2), (5) and (6), coincide if we assume that 1nV −Λ = and f g≡ . 4. References ACZEL, J. 1966. Lectures on functional equations and their applications. Academic Press, New York, London. FEYZIOĞLU, A.K. 1990. A course on algebra. Boğaziçi University Publication, Istanbul. RISTESKI, I.B. 2002. Matrix Method for Solving Linear Complex Vector Functional Equations. International Journal of Mathematics and Mathematical Sciences, 29: 217-238. RISTESKI, I.B. 2004. Solution of Some Nonlinear Complex Vector Functional Equations. African Diaspora Journal of Mathematics, 1: 49-57. RISTESKI, I.B. and COVACHEV, V.C. 2000. On Some General Classes of Partial Linear Complex Vector Functional Equations. Sci. Univ. Tokyo J. Math., 36: 105-146. RISTESKI, I.B. and COVACHEV, V.C. 2001. Complex vector functional equations. World Scientific, New Jersey-London-Singapore-Hong Kong. RISTESKI, I.B., TRENČEVSKI, K.G. and COVACHEV, V.C. 1999. A Simple Functional Operator. New York J. Math., 5: 139-142. RISTESKI, I.B., TRENČEVSKI, K.G. and COVACHEV, V.C. 2000a. Cyclic Complex Vector Functional Equations. Appl. Sci., 2: 13-18. RISTESKI, I.B., TRENČEVSKI, K.G. and COVACHEV, V.C. 2000b. On a linear vector functional equation. In Some Problems of Applied Mathematics. Eds A. Ashyralyev and H.A. Yurtsever. Fatih University Publications, Istanbul. Received 7 October 2003 Accepted 25 April 2004