Applied Mathematics and Computation

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1 Applied Mathematics Computation 219 (2013) Contents lists available at SciVerse ScienceDirect Applied Mathematics Computation journal homepage: One-switch utility functions with annuity payments A.E. Abbas a, J. Chudziak b, a Industrial Enterprise Systems Engineering, University of Illinois at Urbana-Champaign, 104 South Mathews Avenue, Urbana, IL 61801, USA b Department of Mathematics, University of Rzeszów, Rejtana 16 C, Rzeszów, Pol article info abstract Keywords: Invariance One-switch Functional equations Utility theory This paper derives the functional forms of multiattribute utility functions that lead to a maximum of one-switch change in preferences between any two uncertain multiperiod cash flows as the decision maker s wealth increases through constant annuity payments. We derive the general continuous non-constant solutions of the corresponding functional equations. Ó 2013 Elsevier Inc. All rights reserved. 1. Introduction One of the most important steps in decision analysis is determining the decisions maker s utility function [12]. Several authors have discussed this issue have presented methods to assess derive the functional form of a single-attribute utility function based on its risk aversion properties [6,10], or by the change in valuation of a lottery as the decision maker s wealth increases [1,2,7,9]. In particular, Pfanzagl [9] showed that if the decision maker s preferences between any two uncertain uni-period lotteries does not change as the decision maker s initial wealth changes, then he must have either a linear or an exponential utility function. Pfanzagl characterized such utility functions by the functional equation Wðx þ zþ ¼ kðzþwðxþþ lðzþ: Bell [7] further developed this notion introduced the idea of characterizing a utility function based on the maximum number of switches that may occur between any two lotteries as the decision maker s wealth increases. To illustrate, suppose that a decision maker prefers lottery A to lottery B. Now suppose that all outcomes of the lotteries are modified by a shift amount z. If the decision maker s preference between the lotteries does not change for any value of z, then he must have either a linear or an exponential utility function. Thus linear exponential utility functions are 0-switch utility functions. On the other h, if preferences between the two lotteries can change, but can change only once, as we increase z, then the decision maker is said to have a 1-switch utility function. The extension to m-switch utility functions is straightforward; there m is the maximum number of preference changes that can occur as we increase z. Bell [7] characterized the functional forms of m-switch utility functions. Abbas Bell [4] (see also [2]) showed that a one-switch utility function, U, must satisfy the system of functional equations Uðx þ zþ ¼KðzÞUðxÞþMðzÞWðxÞþLðzÞ; Wðx þ zþ ¼kðzÞWðxÞþlðzÞ: In many cases that arise in practice, a decision maker may face multi-period uncertain cash flows. Abbas, Aczél, Chudziak [3] discussed the functional forms of multiattribute utility functions that lead to zero-switch change in preferences between multi-period cash flows when a decision maker s initial wealth increases through an annuity that pays a constant Corresponding author. addresses: aliabbas@ad.uiuc.edu (A.E. Abbas), chudziak@univ.rzeszow.pl (J. Chudziak) /$ - see front matter Ó 2013 Elsevier Inc. All rights reserved.

2 7700 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) amount z every time period. This paper derives the functional forms of multiple attribute utility functions that lead to a maximum of one-switch change in preferences. In particular, we consider one-switch preferences over uncertain n-period cash flows as the decision maker s initial wealth increases. The initial wealth is in the form of an annuity payment that pays an equal amount, z, every period for n successive periods, we consider the solutions of the following system of functional equations: Uðx 1 þ z;...; x n þ zþ ¼KðzÞUðx 1 ;...; x n ÞþMðzÞWðx 1 ;...; x n ÞþLðzÞ; Wðx 1 þ z;...; x n þ zþ ¼kðzÞWðx 1 ;...; x n ÞþlðzÞ: The remainder of this paper is structured as follows: Section 2 presents the problem formulation notation. Section 3 presents several Lemmas preliminary results. Section 4 presents the main results the general continuous nonconstant solutions to the system (1) (2). 2. Problem formulation Assume that D is a non-empty open subset of R n (n P 2), V ðx1 ;...;x nþ :¼ fz 2 Rjðx 1 þ z;...; x n þ zþ 2Dg V D :¼ [ V ðx1 ;...;x nþ; ðx 1 ;...;x nþ2d T :¼ fðx 2 x 1 ;...; x n x 1 Þjðx 1 ;...; x n Þ2Dg, for every ðt 1 ;...; t n 1 Þ2T, [ V ðt 1;...;t n 1 Þ :¼ ðx 1 ;...;x nþ2d;ðx 2 x 1 ;...;x n x 1 Þ¼ðt 1 ;...;t n 1 Þ V ðx1 ;...;x nþ: Furthermore, given a function w : T! R, we set [ V w 0 :¼ V ðx1 ;...;x nþ: ðx 1 ;...;x nþ2d;wðx 2 x 1 ;...;x n x 1 Þ 0 Let us recall that a function a : R! R is said to be additive, provided it satisfies aðx þ yþ ¼aðxÞþaðyÞ for x; y 2 R; a function e : R! R is said to be exponential, provided eðx þ yþ ¼eðxÞeðyÞ for x; y 2 R. It is well known (see e.g. [5]) that every additive function a : R! R continuous at a point has the form aðzþ ¼az for z 2 R with some real constant a. Moreover, every non-zero exponential function e : R! R continuous at a point has the form eðzþ ¼e az for z 2 R with some real constant a.in particular, every non-constant additive or exponential function is non-constant on every interval. We consider the system of functional Eqs. (1) (2) for ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ, where U; W : D! R K; L; M; k; l : V D! R are unknown functions. Eq. (2) has been already solved in [3] under the assumptions that D is open, V ðx1 ;...;xnþ is an interval for every ðx 1 ;...x n Þ2D a function V ðx1 ;...;x nþ 3 z! Wðx 1 þ z;...; x n þ zþ is non-constant for atleast one ðx 1 ;...; x n Þ2D. It is not difficult to check that in fact [3, Theorem 4.3] remains true (with the same proof) if, instead of the openness of D, we assume that, for every ðx 1 ;...; x n Þ2D, the set V ðx1 ;...;xnþ is an open interval. Let us recall that result in such a modified version. Theorem 2.1. Let D be a nonempty subset of R n such that V ðx1 ;...;xnþ is an open interval for every ðx 1 ;...; x n Þ2D. Assume that W : D! R; k; l : V D! R a function given by (3) is non-constant for atleast one ðx 1 ;...; x n Þ2D. Then a triple ðw; k; lþ satisfies Eq. (2) if only if one of the following two conditions holds. (s1) There exist a non-constant additive function a : R! R a function w : T! R such that 8 >< kðzþ ¼1 for z 2 V D ; lðzþ ¼aðzÞ for z 2 V D ; >: Wðx 1 ;...; x n Þ¼wðx 2 x 1 ;...; x n x 1 Þþaðx 1 Þ for ðx 1 ;...; x n Þ2D: (s2) There exist a non-constant exponential function e : R! R, a constant c 2 R a not identically zero function w : T! R such that 8 >< kðzþ ¼eðzÞ for z 2 V w 0 ; lðzþ ¼cð1 kðzþþ for z 2 V D ; >: Wðx 1 ;...; x n Þ¼eðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþc for ðx 1 ;...; x n Þ2D: ð1þ ð2þ ð3þ

3 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) In the present paper we determine all solutions of the system 1,2 under the assumption that a function given by (3) is non-constant for atleast one ðx 1 ;...; x n Þ2D. Therefore, it is clear that Theorem 2.1 will play a crucial role in our considerations. However, we also need to know the solutions of (2), for which a function given by (3) is constant for every ðx 1 ;...; x n Þ2D. Such solutions have been determined in [8], where the following result has been proved. Theorem 2.2. Let D be a nonempty subset of R n such that V ðx1 ;...;xnþ is an open interval for every ðx 1 ;...; x n Þ2D. Assume that W : D! R; k; l : V D! R a function given by (3) is constant for every ðx 1 ;...; x n Þ2D. Then a triple ðw; k; lþ satisfies Eq. (2) if only if there exist a constant d 2 R a function w : T! R such that wðt 1 ;...; t n 1 Þ¼d whenever V ðt 1;...;t n 1 Þ n k 1 ðf1gþ ;; ð4þ lðzþ ¼dð1 kðzþþ for z 2 V D ð5þ Wðx 1 ;...; x n Þ¼wðx 2 x 1 ;...; x n x 1 Þ for ðx 1 ;...; x n Þ2D: ð6þ 3. Preliminaries In this section we assume that D is a nonempty subset of R n such that, for every ðx 1 ;...; x n Þ2D, V ðx1 ;...;xnþ is an open interval. We begin with the following remark. Remark 3.1. Since, for every ðx 1 ;...; x n Þ2D, V ðx1 ;...;xnþ is an open interval containing 0, it is clear that V D ; V ðt 1;...;t n 1 Þ for every ðt 1 ;...; t n 1 Þ2T, V w 0 for every w : T! R, also are open intervals containing 0. Moreover, for every ðt 1 ;...; t n 1 Þ2T; V ðt 1;...;t n 1 Þ is symmetric with respect to 0. In fact, if z 2 V ðt 1;...;t n 1 Þ then z 2 V ðx1 ;...;xnþ for some ðx 1 ;...; x n Þ2D with ðx 2 x 1 ;...; x n x 1 Þ¼ðt 1 ;...; t n 1 Þ. Thus ðx 1 þ z;...; x n þ zþ 2D, whence z 2 V ðx1 þz;...;xnþzþ. As ðx 2 þ z ðx 1 þ zþ;...; x n þ z ðx 1 þ zþþ ¼ ðt 1 ;...; t n 1 Þ, this means that z 2 V ðt 1;...;t n 1 Þ. Next, we prove a series of lemmas, which will play a significant role in the proof of our main result. Lemma 3.1. Let I be an open real interval containing 0 f ; g; F; G : I! R. Assume that G is not identically 0 on every subinterval of I containing 0, for every z 2 I, there exists an open interval V z I containing 0 such that f ðzþgðvþ ¼FðzÞGðvÞ for v 2 V z : ð7þ Then there exists a C 2 R such that FðzÞ ¼Cf ðzþ for z 2 I: ð8þ Proof. Since G is not identically 0 on every subinterval of I containing 0, there exists a sequence ðv r jr 2 NÞ of elements of I such that lim r!1 v r ¼ 0 Gðv r Þ 0 for r 2 N. Note that, for every z 2 I sufficiently large r 2 N, we have v r 2 V z. Thus, by (7), FðzÞ ¼f ðzþ gðv rþ Gðv r Þ ; for z 2 I sufficiently large r 2 N. Therefore, if f is identically 0, (8) holds with any C 2 R. Otherwise, taking z 0 2 I with f ðz 0 Þ 0, we get gðvrþ ¼ Fðz 0Þ for sufficiently large r 2 N which, together with (9), gives (8) with C :¼ Fðz 0Þ. h GðvrÞ f ðz 0 Þ f ðz 0 Þ ð9þ Lemma 3.2. Let A ~ : V D! R. Assume that for every ðx 1 ;...; x n Þ2D; z 2 V ðx1 ;...;xnþ every open interval V z V ðx1 ;...;xnþ containing 0 satisfying z þ V z :¼fz þ vjv 2 V z gv ðx1 ;...;x nþ; ð10þ it holds ~Aðz þ vþ ¼ AðzÞþ ~ AðvÞ ~ for v 2 V z : ð11þ Then there exists a unique additive function A : R! R such that ~AðuÞ ¼AðuÞ for u 2 V D : ð12þ Proof. Fix ðx 1 ;...; x n Þ2D; z 2 V ðx1 ;...;xnþ let V z V ðx1 ;...;xnþ be an open interval containing 0 satisfying (10) (since V ðx1 ;...;xnþ is open, such an interval exists). Moreover, let I z be an open interval containing 0 such that z þ I z V ðx1 ;...;xnþ z þ V z þ I z V ðx1 ;...;xnþ. Then, for every u 2 I z, we have z þ u 2 V ðx1 ;...;xnþ z þ u þ V z V ðx1 ;...;xnþ whence, by the assumptions, we obtain

4 7702 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) ~Aðw þ vþ ¼ ~ AðwÞþ ~ AðvÞ for ðw;vþ 2ðz þ I z ÞV z ; Therefore, applying the result of Radó Baker (1987) [11], we conclude that there is an additive function A z : R! R such that ~AðuÞ ¼A z ðuþ for u 2ðz þ I z Þ[V z : ð13þ Next, taking an arbitrary z 0 2 V ðx1 ;...;xnþ an open interval V z 0 containing 0 such that V z 0 V ðx1 ;...;xnþ z 0 þ V z 0 V ðx1 ;...;xnþ, using the same arguments, we obtain that there exist an open interval I z 0 containing 0 an additive function A z 0 : R! R such that ~AðuÞ ¼A z 0ðuÞ for u 2ðz 0 þ I z 0Þ[V z 0: ð14þ From (13) (14) it follows that A z ðuþ ¼A z 0ðuÞ for u 2 V z \ V z 0. Since the latter set is an open interval (containing 0), this means that A z ¼ A z 0. In this way we have proved there exists an additive function A ðx1 ;...;xnþ : R! R such that ~AðuÞ ¼A ðx1 ;...;x nþðuþ for u 2 V ðx1 ;...;x nþ: ð15þ Now, if ðy 1 ;...; y n Þ2D is arbitrary then, arguing as previously, we obtain that there exists an additive function A ðy1 ;...;y n Þ : R! R such that ~AðuÞ ¼A ðy1 ;...;y n ÞðuÞ for u 2 V ðy1 ;...;y n Þ: Thus, in view of (15), we get A ðx1 ;...;xnþðuþ ¼A ðy1 ;...;y n ÞðuÞ for u 2 V ðx1 ;...;xnþ \ V ðy1 ;...;y n Þ. Since the latter set is an open interval, this yields A ðx1 ;...;xnþ ¼ A ðy1 ;...;y n Þ. Therefore, we have proved that there exists an additive function such that (12) holds. A uniqueness follows from the fact that V D is an interval. h Lemma 3.3. Assume that U; W : D! R K; L; M; k; l : V D! R satisfy the system (1) (2). Let ðx 1 ;...; x n Þ2D; z 2 V ðx1 ;...;xnþ let V z V ðx1 ;...;xnþ be an open interval satisfying (10). Then ½MðzÞðkðvÞ KðvÞÞ MðvÞðkðzÞ KðzÞÞŠWðx 1 ;...; x n Þ¼LðvÞð1 KðzÞÞ LðzÞð1 KðvÞÞ þ MðvÞlðzÞ MðzÞlðvÞ for v 2 V z : ð16þ Proof. In view of (10), for every v 2 V z, we have z 2 V ðx1 þv;...;xnþvþ v 2 V ðx1 þz;...;xnþzþ. Therefore, making use of (1) (2), for every v 2 V z, we obtain Uðx 1 þ v þ z;...; x n þ v þ zþ ¼KðzÞUðx 1 þ v;...; x n þ vþþmðzþwðx 1 þ v;...; x n þ vþþlðzþ; ¼ KðzÞKðvÞUðx 1 ;...; x n ÞþðKðzÞMðvÞþMðzÞkðvÞÞWðx 1 ;...; x n ÞþKðzÞLðvÞþMðzÞlðvÞþLðzÞ Uðx 1 þ z þ v;...; x n þ z þ vþ ¼KðvÞUðx 1 þ z;...; x n þ zþþmðvþwðx 1 þ z;...; x n þ zþþlðvþ; ¼ KðvÞKðzÞUðx 1 ;...; x n ÞþðKðvÞMðzÞþMðvÞkðzÞÞWðx 1 ;...; x n ÞþKðvÞLðzÞþMðvÞlðzÞþLðvÞ: Then a direct computation gives (16). h Lemma 3.4. Assume that U; W : D! R K; L; M; k; l : V D! R satisfy the system (1) (2) M is not identically 0. If the assertion ðs1þ of Theorem 2.1 holds, then there exists a d 2 R such that 1 KðzÞ ¼dMðzÞ for z 2 V D : ð17þ Proof. Assume that ðs1þ holds fix z 2 V D. Let ðx 1 ;...; x n Þ2D be such that z 2 V ðx1 ;...xnþ. Since V ðx1 ;...xnþ is an open interval, there exists an open interval V z containing 0 satisfying (10). Furthermore, using again the openness of V ðx1 ;...xnþ, we can find an open interval I 0 containing 0 such that V z þ I 0 V ðx1 ;...xnþ z þ V z þ I 0 V ðx1 ;...xnþ. Thus, for every u 2 I 0, we have V z V ðx1 þu;...xnþuþ z þ V z V ðx1 þu;...xnþuþ. So, taking into account ðs1þ applying Lemma 3.3 to ðx 1 þ u;...x n þ uþ for u 2 I 0, we conclude that ½MðzÞð1 KðvÞÞ MðvÞð1 KðzÞÞŠWðx 1 þ u;...; x n þ uþ ¼LðvÞð1 KðzÞÞ LðzÞð1 KðvÞÞ þ MðvÞaðzÞ MðzÞaðvÞ for v 2 V z u 2 I 0. On the other h, from ðs1þ it follows that a function given by (3) is non-constant on I 0. Therefore, from the latter equality we derive that MðzÞð1 KðvÞÞ ¼ MðvÞð1 KðzÞÞ for v 2 V z ð18þ

5 LðvÞð1 KðzÞÞ MðzÞaðvÞ ¼LðzÞð1 KðvÞÞ MðvÞaðzÞ for v 2 V z : ð19þ In this way, we have proved that, for every z 2 V D, there is an open interval V z V D such that (18) (19) hold. So, if M is not identically 0 on every subinterval of V D containing 0, then applying Lemma 3.1 (with I ¼ V D )to(18), we obtain that there is a d 2 R such that (17) holds. Suppose that there exists an open interval J V D containing 0 such that MðvÞ ¼0 for v 2 J: ð20þ Since M is not identically 0, taking z 0 2 V D with Mðz 0 Þ 0 applying (18) with z ¼ z 0, we obtain that KðvÞ ¼1 for v 2 J 0 :¼ V z0 \ J: Hence, in view of (19), for every z 2 V D, we get A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) MðzÞaðvÞ ¼LðvÞð1 KðzÞÞ for v 2 V z \ J 0 : ð21þ Moreover a, being a non-constant additive function, is not identically 0 on every subinterval of V D. Thus, applying Lemma 3.1 (with I ¼ V D )to(21), we obtain that there is a C 2 R such that MðzÞ ¼Cð1 KðzÞÞ for z 2 V D. Note also that, as M is not identically 0, we have C 0. Therefore, (17) holds with d :¼ 1. h C Lemma 3.5. Assume that U; W : D! R K; L; M; k; l : V D! R satisfy the system 1,2 M jvw 0 is not identically 0. If the assertion ðs2þ of Theorem 2.1 holds, then there exists a d 2 R such that eðzþ KðzÞ ¼dMðzÞ for z 2 V w 0 : ð22þ Proof. Assume that ðs2þ holds. Let Note that D 0 :¼fðx 1 ;...; x n Þ2Djwðx 2 x 1 ;...; x n x 1 Þ 0g: V D0 ¼ V w 0 ð24þ, by ðs2þ, for every ðx 1 ;...; x n Þ2D 0, a function given by (3) is non-constant on every open subinterval of V ðx1 ;...;xnþ. So, arguing as in the proof of Lemma 3.4, we conclude that, for every z 2 V w 0, there exists an open interval V z V w 0 such that MðzÞðeðvÞ KðvÞÞ ¼ MðvÞðeðzÞ KðzÞÞ for v 2 V z ð25þ LðvÞð1 KðzÞÞ þ cmðvþð1 eðzþþ ¼ LðzÞð1 KðvÞÞ þ cmðzþð1 eðvþþ for v 2 V z : ð26þ If M is not identically 0 on every subinterval of V w 0 containing 0, then applying Lemma 3.1 (with I ¼ V w 0 )to(25), we obtain (22) with some d 2 R. If there exists an open interval J V w 0 containing 0 such that (20) holds, then taking z 0 2 V w 0 with Mðz 0 Þ 0 applying (25) with z ¼ z 0, we get KðvÞ ¼eðvÞ for v 2 J 0 :¼ V z0 \ J: ð27þ Hence, in view of (20) (26), we obtain ð1 KðzÞÞLðvÞ ¼ðLðzÞþcMðzÞÞð1 eðvþþ for z 2 V w 0 ;v 2 V z \ J 0 : ð28þ Furthermore e, being a non-constant exponential function, is not identically 1 on every subinterval of V w 0. Thus, applying Lemma 3.1 (with I ¼ V w 0 )to(28), we conclude that there is a C 2 R such that LðzÞþcMðzÞ ¼Cð1 KðzÞÞ for z 2 V w 0 : ð29þ From (1), (20), (27) (29) we derive that Uðx 1 þ v;...; x n þ vþ ¼eðvÞUðx 1 ;...; x n ÞþCð1 eðvþþ: for every ðx 1 ;...; x n Þ2D 0 v 2 V ðx1 ;...;xnþ \ J 0. Let rðzþ :¼ minfr 2 Nj z 2 J r 0g for z 2 V w 0. Then, in view of (30), for every ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ, we get by induction Uðx 1 þ z;...; x n þ zþ ¼U x 1 þ rðzþ z rðzþ ;...; x n þ rðzþ z rðzþ rðzþ 1 z z X i z ¼¼ e Uðx 1 ;...; x n ÞþC 1 e e : rðzþ rðzþ rðzþ rðzþ i¼0 ð23þ ð30þ As e is exponential, this yields Uðx 1 þ z;...; x n þ zþ ¼eðzÞUðx 1 ;...; x n ÞþCð1 eðzþþ ð31þ

6 7704 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) for ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ. Note that, for every ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ, it holds wðx 2 þ z ðx 1 þ zþ;...; x n þ z ðx 1 þ zþþ ¼ wðx 2 x 1 ;...; x n x 1 Þ 0: So, for every ðx 1 ;...; x n Þ2D 0, we have fz 2 Rjðx 1 þ z;...; x n þ zþ 2D 0 g¼v ðx1 ;...;xnþ. Thus, the set D 0 satisfies the assumptions of Theorems Therefore, if a function V ðx1 ;...;x nþ 3 z! Uðx 1 þ z;...; x n þ zþ is non-constant for atleast one ðx 1 ;...; x n Þ2D 0, applying Theorem 2.1 to (31), we conclude that there exists a not identically zero function W : T 0 :¼fðx 2 x 1 ;...; x n x 1 Þjðx 1 ;...; x n Þ2D 0 g! R such that Uðx 1 ;...; x n Þ¼eðx 1 ÞWðx 2 x 1 ;...; x n x 1 ÞþC for ðx 1 ;...; x n Þ2D 0 : ð34þ Moreover, since e is a non-constant exponential function, it is non-constant on every interval. Hence, for every ðx 1 ;...; x n Þ2D 0, we have V ðx1 ;...;xnþ n e 1 ðf1gþ ;. Thus, in the case where a function given by (33) is constant for every ðx 1 ;...; x n Þ2D 0, applying Theorem 2.2 to (31), we obtain that Uðx 1 ;...; x n Þ¼C for every ðx 1 ;...; x n Þ2D 0. So, again (34) is valid, but this time with W ¼ 0. Next, using again the fact that e is exponential, by (32) (34), we get Uðx 1 þ z;...; x n þ zþ ¼eðx 1 ÞeðzÞWðx 2 x 1 ;...; x n x 1 ÞþC for ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ. On the other h, (1), (34) (s2) imply that Uðx 1 þ z;...; x n þ zþ ¼KðzÞðeðx 1 ÞWðx 2 x 1 ;...; x n x 1 ÞþCÞþMðzÞðeðx 1 Þwðx 2 x 1 ;...; x n x 1 ÞþcÞþLðzÞ for ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ. Hence, taking into account (29), we obtain ðeðzþ KðzÞÞWðx 2 x 1 ;...; x n x 1 Þ¼MðzÞwðx 2 x 1 ;...; x n x 1 Þ for ðx 1 ;...; x n Þ2D 0 z 2 V ðx1 ;...;xnþ. Let b : T 0! R be given by bðt 1 ;...; t n 1 Þ¼ Wðt 1;...; t n 1 Þ wðt 1 ;...; t n 1 Þ for ðt 1 ;...; t n 1 Þ2T 0 : Then from (35) we derive that, for every ðx 1 ;...; x n Þ2D 0, it holds MðzÞ ¼bðx 2 x 1 ;...; x n x 1 ÞðeðzÞ KðzÞÞ for z 2 V ðx 2 x 1 ;...;x n x 1 Þ : ð36þ Since M jvw 0 is not identically 0, in view of (24), there exists ð ~x 1 ;...; ~x n Þ2D 0 such that M jv ð x1 ~ is not identically 0. Moreover we have V ð x1 ~ ;...; xnþ ~ V ð x~ 2 x~ 1 ;...; xn ~ x~ 1 Þ, which means that also M jv ;...; xnþ ~ ð x~ 2 x~ 1 ;...; xn ~ x~ 1 Þ is not identically 0. Thus, taking b :¼ bð ~x 2 ~x 1 ;...; ~x n ~x 1 Þ making use of (36), we obtain that b 0, Kð~zÞ eð~zþ for some ~z 2 V ð x~ 2 x~ 1 ;...; x~ n x~ 1 Þ MðzÞ ¼bðeðzÞ KðzÞÞ for z 2 V ð x~ 2 x~ 1 ;...; x~ n x~ 1 Þ : Now, fix an arbitrary ðx 1 ;...; x n Þ2D 0.IfV ðx 2 x 1 ;...;xn x 1 Þ V ð x~ 2 x~ 1 ;...; xn ~ x~ 1 Þ, then (38) implies that MðzÞ ¼bðeðzÞ KðzÞÞ for z 2 V ðx 2 x 1 ;...;x n x 1 Þ : ð39þ Otherwise, according to Remark 3.1, we have V ð x~ 2 x~ 1 ;...; x~ n x~ 1 Þ V ðx 2 x 1 ;...;x n x 1 Þ ; which, together with (36) (38), gives bðx 2 x 1 ;...; x n x 1 Þ¼b. Thus, by (36), we get again (39). Therefore, for every z 2 S ðx 1 ;...;xnþ2d 0 V ðx 2 x 1 ;...;xn x 1 Þ ¼ V D0, it holds MðzÞ ¼bðeðzÞ KðzÞÞ. So, making use of (24), we obtain (22) with d :¼ 1. h b ð32þ ð33þ ð35þ ð37þ ð38þ 4. Main results Let us begin this section with the following observation. Remark 4.1. Dealing with the solutions of the system 1,2 it is reasonable to assume that M is not identically 0. In fact, if M is identically 0, then equations (1) (2) are independent, each of them can be easily solved by applying Theorems The next theorem is a main result of the paper. Theorem 4.1. Let D be a nonempty subset of R n ðn P 2Þ such that, for every ðx 1 ;...; x n Þ2D; V ðx1 ;...;xnþ is an open interval. Assume that U; W : D! R; K; L; M; k; l : V D! R; M is not identically 0 a function given by (3) is non-constant for atleast one ðx 1 ;...; x n Þ2D. Then the functions U; W; K; L; M; k; l satisfy the system (1) (2) if only if one of the following conditions holds:

7 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) (I) (s1) is valid there exist an additive function A : R! R, a function / : T! R a constant C 2 R such that KðzÞ ¼1 for z 2 V D ; ð40þ MðzÞ ¼CaðzÞ for z 2 V D ; ð41þ LðzÞ ¼AðzÞþ C 2 aðzþ2 for z 2 V D ; ð42þ Uðx 1 ;...; x n Þ¼C aðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþ 1 2 aðx 1Þ 2 þ Aðx 1 Þþ/ðx 2 x 1 ;...; x n x 1 Þ ð43þ (II) (s1) is valid there exist a non-constant exponential function E : R! R, a not identically zero function / : T! R constants b; C 2 R such that KðzÞ ¼EðzÞ for z 2 V / 0 ; ð44þ MðzÞ ¼bð1 KðzÞÞ for z 2 V D ; ð45þ LðzÞ ¼baðzÞþCð1 KðzÞÞ for z 2 V D ; ð46þ Uðx 1 ;...; x n Þ¼b½aðx 1 Þþwðx 2 x 1 ;...; x n x 1 ÞŠ þ Eðx 1 Þ/ðx 2 x 1 ;...; x n x 1 ÞþC ð47þ (III) (s1) is valid there exist a function / : T! R constants b; C 2 R such that (45) (46) hold, /ðt 1 ;...; t n 1 Þ¼C whenever V ðt 1;...;t n 1 Þ n K 1 ðf1gþ ;; ð48þ Uðx 1 ;...; x n Þ¼b½aðx 1 Þþwðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 ÞŠ ð49þ (IV) (s2) is valid there exist a non-constant additive function A : R! R, a function / : T! R a constant b 2 R such that (40) holds, LðzÞ ¼AðzÞ cmðzþ for z 2 V D ; ð50þ MðzÞ ¼bðeðzÞ 1Þ for z 2 V w 0 ; ð51þ Uðx 1 ;...; x n Þ¼beðx 1 Þwðx 2 x 1 ;...; x n x 1 ÞþAðx 1 Þþ/ðx 2 x 1 ;...; x n x 1 Þ ð52þ (V) (s2) is valid there exist a non-constant exponential function E : R! R with E e, a not identically zero function / : T! R constants b; C 2 R such that (44) holds, LðzÞ ¼Cð1 KðzÞÞ cmðzþ for z 2 V D ; ð53þ MðzÞ ¼bðeðzÞ EðzÞÞ for z 2 V w 0 ; ð54þ Uðx 1 ;...; x n Þ¼bðeðx 1 Þwðx 2 x 1 ;...; x n x 1 ÞþcÞþEðx 1 Þ/ðx 2 x 1 ;...; x n x 1 ÞþC ð55þ (VI) (s2) is valid there exist a function / : T! R constants b; C 2 R such that (48) (53) hold, MðzÞ ¼bðeðzÞ KðzÞÞ for z 2 V w 0 ; ð56þ Uðx 1 ;...; x n Þ¼beðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 Þ ð57þ (VII) (s2) is valid, there exist an additive function A : R! R, a function / : T! R a constant C 2 R such that (53) holds, KðzÞ ¼eðzÞ for z 2 V w 0 ; ð58þ MðzÞ ¼eðzÞAðzÞ for z 2 V w 0 ; ð59þ Uðx 1 ;...; x n Þ¼eðx 1 Þ½Aðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 ÞŠ þ C for ðx 1 ;...; x n Þ2D: ð60þ Proof. Assume that the functions U; W; K; L; M; k; l satisfy the system 1,2. Then, according to Theorem 2.1, one of the conditions ðs1þ ðs2þ is satisfied. First consider the situation, where ðs1þ holds. Then, in view of Lemma 3.4, we get (17) with some d 2 R. It will be convenient to distinguish the following two cases:

8 7706 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) d 0; 2. d ¼ 0. Case 1. From (1), (2) (17) it follows that ðw duþðx 1 þ z;...; x n þ zþ ¼KðzÞðW duþðx 1 ;...; x n ÞþlðzÞ dlðzþ; for ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Furthermore, since d 0 M is not identically 0, by (17), we get that K is not identically 1. Therefore, applying Theorems 2.1,2.2 to (61), we obtain that either there exist a non-constant exponential function E : R! R, a not identically zero function ~ / : T! R a constant ~ C 2 R such that KðzÞ ¼EðzÞ for z 2 V ~/ 0 ; ð62þ lðzþ dlðzþ ¼ ~ Cð1 KðzÞÞ for z 2 V D ð63þ ðw duþðx 1 ;...; x n Þ¼Eðx 1 Þ ~ /ðx 2 x 1 ;...; x n x 1 Þþ ~ C ð64þ or there exist a function ~ / : T! R a constant ~ C 2 R such that (63) holds, ~/ðt 1 ;...; t n 1 Þ¼ ~ C whenever V ðt 1;...;t n 1 Þ n K 1 ðf1gþ ; ð65þ ðw duþðx 1 ;...; x n Þ¼ ~ /ðx 2 x 1 ;...; x n x 1 Þ for ðx 1 ;...; x n Þ2D: ð66þ Hence, taking b :¼ 1 d ; C :¼ ~ C d / :¼ 1 d ~ /, in view of (17) ðs1þ, we obtain ðiiþ ðiiiþ, respectively. Case 2. We prove that in this case, ðiþ holds. Since d ¼ 0, (17) implies (40). Fix ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Let V z V ðx1;...;xnþ be an arbitrary open interval containing 0 2 V z satisfying (10). Then, according to Lemma 3.3, (16) holds, whence making use of (40) ðs1þ, we get MðzÞaðvÞ ¼MðvÞaðzÞ for v 2 V z : ð67þ Furthermore, since a is a non-constant additive function, it is not identically 0 on every interval. Therefore, applying Lemma 3.1 (with I ¼ V D )to(67), we obtain (41) with some C 2 R. Next, in view of (1), (2), (40), (41)ðs1Þ, for every v 2 V z, we get Hence Lðz þ vþþcaðz þ vþwðx 1 ;...x n ÞþUðx 1 ;...x n Þ¼Lðz þ vþþmðz þ vþwðx 1 ;...x n ÞþKðz þ vþuðx 1 ;...x n Þ ¼ Uðx 1 þ z þ v;...x n þ z þ vþ ¼ KðzÞUðx 1 þ v;...x n þ vþþmðzþwðx 1 þ v;...x n þ vþþlðzþ ¼ KðzÞðKðvÞUðx 1 ;...x n ÞþMðvÞWðx 1 ;...x n ÞþLðvÞÞ þ MðzÞ ðkðvþwðx 1 ;...x n ÞþlðvÞÞ þ LðzÞ ¼ KðzÞKðvÞUðx 1 ;...x n ÞþðMðzÞkðvÞþMðvÞÞWðx 1 ;...x n ÞþKðzÞLðvÞ þ MðzÞlðvÞ þlðzþ ¼ Uðx 1 ;...x n ÞþC½aðzÞþaðvÞŠWðx 1 ;...x n ÞþLðvÞþCaðzÞaðvÞþLðzÞ: Lðz þ vþ ¼LðzÞþLðvÞþCaðzÞaðvÞ for v 2 V z : ð68þ Let a function ~ A : V D! R be given by ~AðuÞ :¼ LðuÞ C 2 aðuþ2 for u 2 V D : ð61þ Then, by (68), we get (11). Hence, applying Lemma 3.2, we conclude that there exists an additive function A : R! R such that (12) holds. Therefore, (42) is valid. Next, define a function U ~ : D! R by ~Uðx 1 ;...; x n Þ¼Uðx 1 ;...; x n Þ C aðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþ 1 2 aðx 1Þ 2 ð69þ for ðx 1 ;...; x n Þ2D. Then, making use of (1) (40) (42) after straightforward calculations, we obtain that ~Uðx 1 þ z;...; x n þ zþ ¼ ~ Uðx 1 ;...; x n ÞþAðzÞ for ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Hence, applying Theorem 2.1 in the case where A is not identically 0; Theorem 2.2, otherwise, we conclude that there exists a function / : T! R such that

9 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) ~Uðx 1 ;...; x n Þ¼/ðx 2 x 1 ;...; x n x 1 ÞþAðx 1 Þ for ðx 1 ;...; x n Þ2D: Thus, taking into account (69), we get (43) so ðiþ holds. Now, we are going to deal with the situation, where ðs2þ is valid. First consider the case, where K ¼ k on V w 0, that is (58) holds. We show that in this case ðviiþ is valid. Fix ðx 1 ;...; x n Þ2D 0 (see (23)). Then wðx 2 x 1 ;...; x n x 1 Þ 0, whence Wðx 1 ;...x n Þ c. Let z 2 V ðx1 ;...;xnþ let V z V ðx1 ;...;xnþ be an arbitrary open interval containing 0 satisfying (10). Then, according to Lemma 3.3, (16) holds. Thus, taking into account (58) ðs2þ, we obtain ðlðvþþcmðvþþð1 eðzþþ ¼ ðlðzþþcmðzþþð1 eðvþþ for v 2 V z : Moreover e, being a non-constant exponential function, is not identically 1 on every interval. So, applying Lemma 3.1 (with I ¼ V w 0 ), we get that there is a C 2 R such that LðzÞþcMðzÞ ¼Cð1 eðzþþ for z 2 V w 0 : ð70þ Next, making use of (1), (2), (10) (58), for every v 2 V z, we obtain eðz þ vþuðx 1 ;...x n ÞþMðz þ vþwðx 1 ;...x n ÞþLðz þ vþ ¼Uðx 1 þ z þ v;...x n þ z þ vþ ¼ eðvþuðx 1 þ z;...x n þ zþþmðvþwðx 1 þ z;...x n þ zþþlðvþ ¼ eðvþeðzþuðx 1 ;...x n Þþ½eðvÞMðzÞþMðvÞeðzÞŠWðx 1 ;...x n Þ þ eðvþlðzþþmðvþlðzþ þlðvþ: Since e is exponential, this yields ½Mðz þ vþ MðzÞeðvÞ MðvÞeðzÞŠWðx 1 ;...x n Þ¼eðvÞLðzÞþMðvÞlðzÞþLðvÞ Lðz þ vþ for v 2 V z : On the other h, from ðs2þ (70) it follows that eðvþlðzþþmðvþlðzþþlðvþ Lðz þ vþ ¼eðvÞðCð1 eðzþþ cmðzþþ þ cmðvþð1 eðzþþ þ Cð1 eðvþþ cmðvþ Cð1 eðz þ vþþ þ cmðz þ vþ ¼cðMðz þ vþ MðzÞeðvÞ MðvÞeðzÞÞ for v 2 V z : Therefore, we get ½Mðz þ vþ MðzÞeðvÞ MðvÞeðzÞŠðWðx 1 ;...x n Þ cþ ¼0 for v 2 V z : Since Wðx 1 ;...x n Þ c, this implies that is Mðz þ vþ ¼MðzÞeðvÞþMðvÞeðzÞ for v 2 V z ; Mðz þ vþ eðz þ vþ ¼ MðzÞ eðzþ þ MðvÞ eðvþ for v 2 V z : Thus, applying Lemma 3.2 (with I ¼ V w 0 A ~ ¼ M ), we conclude that there exists an additive function A : R! R such e that MðzÞ ¼ AðzÞ for z 2 V eðzþ w 0. This yields (59). Next, define a function U ~ : D! R by ~Uðx 1 ;...; x n Þ¼Uðx 1 ;...; x n Þ eðx 1 ÞAðx 1 Þwðx 2 x 1 ;...; x n x 1 Þ for ðx 1 ;...; x n Þ2D. We claim that ð71þ ~Uðx 1 þ z;...; x n þ zþ ¼KðzÞ ~ Uðx 1 ;...; x n ÞþcMðzÞþLðzÞ for ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Fix ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Ifz 2 V w 0, then making use of (1, 58, 59, 70, 71)ðs2Þ, we obtain ~Uðx 1 þ z;...; x n þ zþ ¼Uðx 1 þ z;...; x n þ zþ eðx 1 þ zþaðx 1 þ zþwðx 2 x 1 ;...; x n x 1 Þ ¼ KðzÞUðx 1 ;...; x n ÞþMðzÞWðx 1 ;...; x n ÞþLðzÞ eðx 1 þ zþaðx 1 þ zþwðx 2 x 1 ;...; x n x 1 Þ ¼ eðzþuðx 1 ;...; x n ÞþeðzÞAðzÞðeðx 1 Þwðx 2 x 1 ;...; x n x 1 ÞþcÞ þ LðzÞ eðx 1 ÞeðzÞAðx 1 Þwðx 2 x 1 ;...; x n x 1 Þ eðx 1 ÞeðzÞAðzÞwðx 2 x 1 ;...; x n x 1 Þ ¼ eðzþðuðx 1 ;...; x n Þ eðx 1 ÞAðx 1 Þwðx 2 x 1 ;...; x n x 1 ÞÞþceðzÞAðzÞþLðzÞ ¼ KðzÞ Uðx ~ 1 ;...; x n ÞþcMðzÞþLðzÞ: If z R V w 0, then wðx 2 x 1 ;...; x n x 1 Þ¼0, whence Wðx 1 ;...; x n Þ¼c, ~ Uðx1 ;...; x n Þ¼Uðx 1 ;...; x n Þ ~Uðx 1 þ z;...; x n þ zþ ¼Uðx 1 þ z;...; x n þ zþ. Thus, (72) follows directly from (1). Therefore we have shown that (72) holds for every ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Since, by (58), K is not identically 1 on every interval, applying Theorems ,to(72), we obtain that either there exist a non-constant exponential function ~e : R! R, a not identically zero function / : T! R a constant ~ C 2 R such that ð72þ

10 7708 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) KðzÞ ¼~eðzÞ for z 2 V / 0 ; ð73þ LðzÞþcMðzÞ ¼ ~ Cð1 KðzÞÞ for z 2 V D ð74þ ~Uðx 1 ;...; x n Þ¼~eðx 1 Þ/ðx 2 x 1 ;...; x n x 1 Þþ ~ C ð75þ or there exists a constant ~ C 2 R such that (74) holds ~ Uðx 1 ;...; x n Þ¼ ~ C for ðx 1 ;...; x n Þ2D. In the first case, by (58) (73), we get eðzþ ¼~eðzÞ for z 2 V w 0 \ V / 0. Since the latter set is an open interval, this yields that e ¼ ~e. Furthermore, from (58), (70) (74), we deduce that C ¼ ~ C. Thus, (74) implies (53) (75), together with (71), gives (60). Therefore, ðviiþ is valid. In the second case, we obtain ðviiþ with C ¼ ~ C / ¼ 0. Finally, consider the case where Kðz 0 Þ eðz 0 Þ for some z 0 2 V w 0 : ð76þ If M jvw 0 is identically 0, then MðzÞWðx 1 ;...; x n Þ¼cMðzÞ for every ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ, whence (1) becomes Uðx 1 þ z;...x n þ zþ ¼KðzÞUðx 1 ;...x n ÞþcMðzÞþLðzÞ for ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ. Therefore, applying Theorems , we obtain ðivþ ðviþ with b ¼ 0 (note that (76) implies E e in ðvþ). Assume that M jvw 0 is not identically 0. Then, according to Lemma 3.5, (22) holds with some d 2 R. Furthermore, taking into account (76), we get d 0. We claim that, for every ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ, it holds ðw duþðx 1 þ z;...; x n þ zþ ¼KðzÞðW duþðx 1 ;...; x n Þþcð1 KðzÞÞ dðcmðzþþlðzþþ: Fix ðx 1 ;...; x n Þ2D z 2 V ðx1 ;...;xnþ.ifz 2 V w 0 then (1), (2) (22) imply (61). On the other h, making use of (22) ðs2þ, we obtain lðzþ dlðzþ ¼cð1 eðzþþ dlðzþ ¼cð1 KðzÞÞ cðeðzþ KðzÞÞ dlðzþ ¼cð1 KðzÞÞ dðcmðzþþlðzþþ; which together with (61), gives (77). Ifz R V w 0, then Wðx 1 ;...; x n Þ¼Wðx 1 þ z;...; x n þ zþ ¼c so, in view of (1), we get ðw duþðx 1 þ z;...; x n þ zþ ¼c dðkðzþuðx 1 ;...; x n ÞþcMðzÞþLðzÞÞ ¼ KðzÞðc duðx 1 ;...; x n ÞÞ þ cð1 KðzÞÞ dðcmðzþþlðzþþ ¼ KðzÞðW duþðx 1 ;...; x n Þþcð1 KðzÞÞ dðcmðzþþlðzþþ: Thus we obtain again (77). Therefore, applying Theorems 2.1,2.2 to (77), we conclude that the following three cases are possible: ðaþ KðzÞ ¼1 for z 2 V D there exist a non-constant additive function ~ A : R! R a function ~ / : T! R such that ð77þ dðcmðzþþlðzþþ ¼ cð1 KðzÞÞ dðcmðzþþlðzþþ ¼ ~ AðzÞ for z 2 V D ðw duþðx 1 ;...; x n Þ¼ ~ /ðx 2 x 1 ;...; x n x 1 Þþ ~ Aðx 1 Þ ðbþ there exist a non-constant exponential function E : R! R, a not identically zero function ~ / : T! R a constant ~C 2 R such that (62) (64) hold, cð1 KðzÞÞ dðcmðzþþlðzþþ ¼ ~ Cð1 KðzÞÞ for z 2 V D ; ð78þ ðcþ there exist a function ~ / : T! R a constant ~ C 2 R such that (65), (66) (78) hold. Moreover, by (76), in the case of ðbþ, we have E e. Therefore, making use of (22), we obtain ðivþ with A :¼ 1 A; ~ / :¼ 1 d d ðc /Þ, ~ b :¼ 1 d ; ðvþ with / :¼ 1 /; ~ b :¼ 1 d d ; C :¼ ~ C d ; ðviþ with / :¼ 1 d ðc /Þ; ~ b :¼ 1 d ; C :¼ c ~ C d, respectively. Since a converse is easy to check, this completes the proof. h Finally, we present a result describing the solutions of the system 1,2 such that the functions given by (3) (33) are continuous for every ðx 1 ;...; x n Þ2D. Theorem 4.2. Let D be a nonempty subset of R n such that V ðx1 ;...;xnþ is an open interval for every ðx 1 ;...; x n Þ2D. Assume that U; W : D! R; K; L; M; k; l : V D! R; M is not identically 0, the functions given by (3) (33) are continuous for every ðx 1 ;...; x n Þ2D, a function given by (3) is non-constant for atleast one ðx 1 ;...; x n Þ2D. Then the functions U; W; K; L; M; k; l satisfy the system (1) (2) if only if one of the following conditions holds: ðc I Þ there exist functions /; w : T! R constants a 2 R nf0g; A; p 2 R such that (40) holds,

11 kðzþ ¼1 for z 2 V D ; ð79þ lðzþ ¼az for z 2 V D ; ð80þ MðzÞ ¼pz for z 2 V D ; LðzÞ ¼Az þ pa 2 z2 for z 2 V D ; Wðx 1 ;...; x n Þ¼ax 1 þ wðx 2 x 1 ;...; x n x 1 Þ ð81þ Uðx 1 ;...; x n Þ¼ pa 2 x2 1 þ½pwðx 2 x 1 ;...; x n x 1 ÞþAŠx 1 þ /ðx 2 x 1 ;...; x n x 1 Þ ðc II Þ there exist a not identically zero function / : T! R, a function w : T! R constants a; a 2 R nf0g,b; C 2 R such that (45) (79) (81) hold, KðzÞ ¼e az for z 2 V / 0 ; ð82þ LðzÞ ¼abz þ Cð1 KðzÞÞ for z 2 V D ; ð83þ Uðx 1 ;...; x n Þ¼abx 1 þ e ax 1 /ðx 2 x 1 ;...; x n x 1 Þþbwðx 2 x 1 ;...; x n x 1 ÞþC ðc III Þ there exist functions /; w : T! R constants a 2 R nf0g; b; C 2 R such that (45), (48), (79) (81) (83) hold, Uðx 1 ;...; x n Þ¼abx 1 þ b½wðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 ÞŠ ðc IV Þ there exist a function / : T! R, a not identically zero function w : T! R constants A; b 2 R nf0g, b; c 2 R such that (40) holds, kðzþ ¼e bz for z 2 V w 0 ; ð84þ lðzþ ¼cð1 kðzþþ for z 2 V D ; ð85þ LðzÞ ¼Az cmðzþ for z 2 V D ; MðzÞ ¼bðe bz 1Þ for z 2 V w 0 ; Wðx 1 ;...; x n Þ¼e bx 1 wðx 2 x 1 ;...; x n x 1 Þþc ð86þ Uðx 1 ;...; x n Þ¼Ax 1 þ be bx 1 wðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 Þ ðc V Þ there exist not identically zero functions /; w : T! R constants a; b 2 R nf0g, b; c; C 2 R such that a b, (53), (82) (84) (86) hold, MðzÞ ¼bðe bz e az Þ for z 2 V w 0 ; Uðx 1 ;...; x n Þ¼e ax 1 /ðx 2 x 1 ;...; x n x 1 Þþbe bx 1 wðx 2 x 1 ;...; x n x 1 Þþbc þ C ðc VI Þ there exist a function / : T! R, a not identically zero function w : T! R constants b 2 R nf0g, b; C 2 R such that (48), (53), (84) (86) hold, MðzÞ ¼bðe bz KðzÞÞ for z 2 V w 0 ; Uðx 1 ;...; x n Þ¼be bx 1 wðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 Þ ðc VII Þ there exist a function / : T! R, a not identically zero function w : T! R constants b 2 R nf0g, A; C; c 2 R such that (53) (84) (86) hold, KðzÞ ¼e bz for z 2 V w 0 ; A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) MðzÞ ¼Aze bz for z 2 V w 0 ; Uðx 1 ;...; x n Þ¼e bx 1 ½Ax 1 wðx 2 x 1 ;...; x n x 1 Þþ/ðx 2 x 1 ;...; x n x 1 ÞŠ þ C for ðx 1 ;...; x n Þ2D:

12 7710 A.E. Abbas, J. Chudziak / Applied Mathematics Computation 219 (2013) Proof. Assume that the functions U; W; K; L; M; k; l satisfy the system (1) (2). Then, one of the conditions ðiþ ðviiþ of Theorem 4.1 holds. In the case of ðiþ, taking ðx 1 ;...; x n Þ2D an arbitrary sequence ðz r jr 2 NÞ of real numbers such that lim r!1 z r ¼ 0, we have z r 2 V ðx1 ;...;xnþ for sufficiently large r 2 N, whence lim aðx 1 þ z r Þ¼ lim ½Wðx 1 þ z r ;...; x n þ z r Þ wðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞŠ r!1 r!1 ¼ Wðx 1 ;...; x n Þ wðx 2 x 1 ;...; x n x 1 Þ¼aðx 1 Þ lim Aðx 1 þ z r Þ¼ lim fuðx 1 þ z r ;...; x n þ z r Þ C aðx 1 þ z r Þwðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞ þ 1 r!1 r!1 2 aðx 1 þ z r Þ 2 /ðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞg ¼ Uðx 1 ;...; x n Þ C aðx 1 Þwðx 2 x 1 ;...; x n x 1 Þþ 1 2 aðx 1 þ z r Þ 2 /ðx 2 x 1 ;...; x n x 1 Þ¼Aðx 1 Þ: Thus, a A are additive functions continuous at a point x 1. Hence there exist real constants a A such that aðzþ ¼az for z 2 R AðzÞ ¼Az for z 2 R. Moreover, as a is non-constant, we have a 0. Therefore, taking p :¼ Ca, we obtain ðc I Þ. In the case of ðiiþ, taking ðx 1 ;...; x n Þ2D with wðx 2 x 1 ;...; x n x 1 Þ 0, an arbitrary sequence ðz r jr 2 NÞ of real numbers such that lim r!1 z r ¼ 0, as previously, we get lim r!1 aðx 1 þ z r Þ¼aðx 1 Þ lim Eðx Uðx 1 þ z r ;...; x n þ z r Þ 1 þ z r Þ¼lim r!1 r!1 /ðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞ lim r!1 b½aðx 1 þ z r Þþwðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞŠ þ C /ðx 2 þ z r ðx 1 þ z r Þ;...; x n þ z r ðx 1 þ z r ÞÞ ¼ Uðx 1;...; x n Þ b½aðx 1 Þþwðx 2 x 1 ;...; x n x 1 ÞŠ C ¼ Eðx 1 Þ: /ðx 2 x 1 ;...; x n x 1 Þ Thus a is a non-constant additive function continuous at x 1, E is a non-constant exponential function continuous at x 1. So, there exist a; a 2 R nf0g such that aðzþ ¼az for z 2 R EðzÞ ¼e az for z 2 R. Therefore ðc II Þ holds. Using the similar arguments, from ðiiiþ ðviiþ one can easily derive ðc III Þ ðc VII Þ, respectively. h 5. Conclusion The concept of one-switch change in preferences between uncertain lotteries, introduced by Bell [7], is an intuitively appealing concept for thinking about von Neumann Morgenstern utility functions. It also significantly constrains the allowable functional forms thereby facilitating the assessment of a decision maker s utility function. We extended this one-switch concept to situations involving multiple time periods where the initial wealth increases in the form of a constant annuity payment. We derived the corresponding functional forms of the utility functions showed that theses functional forms are also highly constrained, though a bit more in number than the single period case. Our goal was to derive the general continuous non-constant solutions of the system of equations (1) (2) without assumption of differentiability. Eq. (2) alone has been already solved in Abbas, Aczél, Chudziak [3] Chudziak [8]. Our work forms the basis for more specific future formulations that could be used to derive the functional form of a multiattribute utility function to help with utility elicitation in practice, further constrain the functional form. In particular, future work can consider one-switch preferences when the annuity payments are adjusted for inflation, or when the corresponding multiattribute utility function allows for constant discounting. These other formulations, further constrain the allowable functional forms, building on the more general solutions we derived in this work. References [1] A.E. Abbas, Invariant utility functions certain equivalent transformations, Decis. Anal. 4 (2007) [2] A.E. Abbas, J. Aczél, The role of some functional equations in decision analysis, Decis. Anal. 7 (2010) [3] A.E. Abbas, J. Aczél, J. Chudziak, Invariance of multiattribute utility functions under shift transformations, Results. Math. 54 (2009) [4] A.E. Abbas, D. Bell, One- switch conditions for multiattribute utility functions, Oper. Res. 60 (2012) [5] J. Aczél, J. Dhombres, Functional equations in several variables, Encyclopedia of Mathematics its Applications, 31, Cambridge University Press, [6] K. Arrow, The theory of risk aversion, Lecture 2, Aspects of the Theory of Risk-Bearing, Yrjo Jahnssonin Saatio, Helsinki, [7] D. Bell, One-switch utility functions a measure of risk, Manage. Sci. 34 (1988) [8] J. Chudziak, On a class of multiattribute utility functions invariant under shift transformations, Acta Phys. Pol. A 117 (2010) [9] J. Pfanzagl, A general theory of measurement applications to utility, Naval Res. Logist. Q. 6 (1959) [10] J. Pratt, Risk aversion in the small in the large, Econometrica 32 (1964) [11] F. Radó, J.A. Baker, Pexider s equation aggregation of allocations, Aequationes Math. 32 (1987) [12] J. von Neumann, O. Morgenstern, Theory of Games Economic Behavior, Princeton University Press, Princeton, NJ, 1947.